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## Case Study Questions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

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Here we are providing Case Study questions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities.

## Algebraic Expressions and Identities Case Study Questions

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## Unit 8: Algebraic expressions and identities

Addition and subtraction.

- Adding polynomials (Opens a modal)
- Subtracting polynomials (Opens a modal)
- Add polynomials (intro) Get 3 of 4 questions to level up!
- Subtract polynomials (intro) Get 3 of 4 questions to level up!

## Multiplying monomials by monomials

- Multiplying monomials (Opens a modal)
- Multiplying monomials challenge (Opens a modal)
- Multiply monomials Get 3 of 4 questions to level up!
- Multiply monomials (advanced) Get 3 of 4 questions to level up!

## Multiplying monomials by polynomials

- Multiplying monomials by polynomials (Opens a modal)
- Multiplying monomials by polynomials challenge (Opens a modal)
- Multiply monomials by polynomials Get 3 of 4 questions to level up!
- Multiply monomials by polynomials challenge Get 3 of 4 questions to level up!

## Multiplying binomials

- Multiplying binomials intro (Opens a modal)
- Multiplying binomials by polynomials (Opens a modal)
- Multiply binomials intro Get 3 of 4 questions to level up!
- Multiply binomials Get 3 of 4 questions to level up!
- Multiply binomials by polynomials Get 3 of 4 questions to level up!

## Standard identities

- Special products of the form (x+a)(x-a) (Opens a modal)
- Squaring binomials of the form (x+a)² (Opens a modal)
- Squaring binomials of the form (ax+b)² (Opens a modal)
- Multiply difference of squares Get 3 of 4 questions to level up!
- Polynomial special products: perfect square Get 3 of 4 questions to level up!

## Polynomials word problems

- Polynomial word problem: total value of bills (Opens a modal)
- Polynomial word problem: area of a window (Opens a modal)
- Polynomial word problem: rectangle and circle area (Opens a modal)

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## Chapter 8 Class 8 Algebraic Expressions and Identities

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Updated for new NCERT.

Get NCERT Solutions of Chapter 8 Class 8 Algebraic Expressions and Identities free at Teachoo. Answers to all exercise questions, examples have been solved with step-by-step solutions. Concepts are explained before doing the questions.

In this chapter, we will learn

- What are algebra expressions
- Terms, Factors and Coefficients in an Algebra Expression
- What are monomials, binomials, trinomials and polynomials
- What are like and unlike terms in an algebraic expression
- Adding and Subtracting Algebra Expression
- Multiplying two monomials
- Multiplying three or more monomials
- Multiplying Monomial by a Binomial
- Multiplying Monomial by a Trionmial
- Multiplying Binomial by a Binomial
- Multiplying Binomial by a Trionmial
- Algebra Identities

Here, we have divided the chapter into 2 parts - Serial Order Wise and Concept Wise.

Just like the NCERT Book, in Serial Order Wise, the chapter is divided into exercises and examples. This is useful if you are looking for answer to a specific question.

That is not a good way of studying.

In the NCERT Book, first question is of some topic, second question is of some other topic. There is no order

We have solved that using Concept Wise.

In Concept Wise , the chapter is divided into concepts. First the concept is taught. Then the questions of that concept is answered, from easy to difficult.

Click on a link to start doing the chapter

## Serial order wise

Concept wise.

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## NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities

The NCERT solutions for Class 8 maths Chapter 9 Algebraic Expressions and Identities provides an introduction to a useful mathematical tool to handle the ever changing world we live in. If there is one thing that is constant about the universe, it is ‘change’ so it was felt early on by the mathematicians to have a framework such that it can handle the change and everytime the calculation need not be done from scratch. For example, when they studied geometry and measured the area of one circle thereafter using mathematics they devised a formula into which everytime you feed into the value of the radius of the circle it would give the area of the corresponding circle.

This power to have an expression is such that based on the value of the input it will give different answers, thus greatly enhancing the usage of mathematics . Let us take an example in the NCERT solutions Class 8 maths Chapter 9. Consider that in a school , for students of class 7, one notebook per student is needed and for students of class 8 two notebooks per student is needed. If we consider the number of students in class 7 as ‘x’ and number of students in class 8 as ‘y’ then the number of books needed in total would be ‘x + 2y’. The Class 8 maths NCERT solutions Chapter 9 Algebraic Expressions and Identities will help students understand both the utility and the application of such expressions and also you can find some of these in the exercises given below.

- NCERT Solutions Class 8 Maths Chapter 9 Ex 9.1
- NCERT Solutions Class 8 Maths Chapter 9 Ex 9.2
- NCERT Solutions Class 8 Maths Chapter 9 Ex 9.3
- NCERT Solutions Class 8 Maths Chapter 9 Ex 9.4
- NCERT Solutions Class 8 Maths Chapter 9 Ex 9.5

## NCERT Solutions for Class 8 Maths Chapter 9 PDF

Algebraic Expressions and Identities is an important and interesting concept to understand the nature of mathematics, and the NCERT solutions Class 8 maths Chapter 9 Algebraic Expressions and Identities has made sure to analyze this topic step by step, covering the easy examples and exercise questions first, then gradually moving on to the complicated ones so that the students feel motivated throughout their learning. The exercise questions in each segment of the chapter can be accessed through the links below :

## ☛ Download Class 8 Maths NCERT Solutions Chapter 9

NCERT Class 8 Maths Chapter 9 Download PDF

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

Algebraic expressions play an important role in giving the problems an easy frame that can be understood properly and then solved accordingly. Hence, the study of algebra holds an important place in mathematics. A brief classification of questions in each exercise of NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities can be seen as below:

- Class 8 Maths Chapter 9 Ex 9.1 - 4 Questions
- Class 8 Maths Chapter 9 Ex 9.2 - 5 Questions
- Class 8 Maths Chapter 9 Ex 9.3 - 5 Questions
- Class 8 Maths Chapter 9 Ex 9.4 - 3 Questions
- Class 8 Maths Chapter 9 Ex 9.5 - 8 Questions

☛ Download Class 8 Maths Chapter 9 NCERT Book

Topics Covered: The Class 8 math NCERT solutions Chapter 9 covers the concept of coefficients and variables, factors , terms, polynomials , the addition of polynomials , multiplication of polynomials , and algebraic identities .

Total Questions: Class 8 maths Chapter 9 Algebraic Expressions and Identities consists of 25 questions out of which 17 are easy and quick to solve, while the remaining 8 may require a longer time.

## List of Formulas in NCERT Solutions Class 8 Maths Chapter 9

The NCERT solutions Class 8 maths Chapter 9 outlines what an expression is, how it is made and what are the different parts in an expression. Further, how to perform operations on the expressions like those of addition , multiplication , etc., and the explanation of some standard identities is drafted in this chapter. Some of the important facts and formulas covered in NCERT solutions for Class 8 maths Chapter 9 are given below :

- Expressions are made up of terms which in turn are made of coefficients and variables.
- Operations involving expressions follow the distributive law.
- Below are the three identities which hold true for any value of variable
- (a + b) 2 = a 2 + 2ab + b 2
- (a – b) 2 = a 2 – 2ab + b 2
- (a + b) (a – b) = a 2 – b 2

## Important Questions for Class 8 Maths NCERT Solutions Chapter 9

Ncert solutions for class 8 maths video chapter 9, faqs on ncert solutions class 8 maths chapter 9, why are ncert solutions class 8 maths chapter 9 important.

The NCERT Solutions Class 8 Maths Chapter 9 has well-drafted examples to explain the concept of algebraic expressions and identities. The components that make an expression have been dealt with in detail to bring better clarity to the subject and instilling the necessary confidence in the students. The CBSE board asks the students to refer to the NCERT books which is proof of their exceptional quality.

## Do I Need to Practice all the Questions Provided in NCERT Solutions Class 8 Maths Algebraic Expressions and Identities?

Expressions are like mathematical entities of their own and they have widespread applications. This chapter provides the opportunity to learn about this very important and useful concept. Additionally, expressions can vary from simple to complex, and while the concept is simple, its applications can be amazingly creative. Owing to these reasons, one must make sure to study the application of algebraic expressions. With plenty of examples and exercise questions in NCERT Solutions Class 8 Maths Algebraic Expressions and Identities, the students can explore every basic detail.

## What are the Important Topics Covered in NCERT Solutions Class 8 Maths Chapter 9?

The NCERT Solutions Class 8 Maths Chapter 9 covers the concept of Algebraic Expressions and Identities in detail with the help of proper explanations and most importantly by introducing the components which make up an expression that is the ‘terms’. It further zeroes on what makes up a term” that is the ‘coefficient' and ‘variable’, then takes a macroscopic view of how expressions can be of different types and how they operate.

## How Many Questions are there in NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities?

The NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities has 25 questions, 17 of which are short-form, while 8 are long-form questions. The students will be required to calculate or identify Algebraic Expressions and Identities of numbers.

## How CBSE Students can utilize NCERT Solutions Class 8 Maths Chapter 9 effectively?

The students should pay attention to the highlight sections in the NCERT Solutions Class 8 Maths Chapter 9 as they speak about important facts. These should be noted down and revised along with a consistent practice of solved examples and exercise questions. In this way, the students can utilize the NCERT Solutions Class 8 Maths Chapter 9 effectively.

## Why Should I Practice NCERT Solutions Class 8 Maths Algebraic Expressions and Identities Chapter 9?

The NCERT Solutions Class 8 Maths Algebraic Expressions and Identities Chapter 9 has well-explained examples and a logical step-by-step explanation of the facts with proper problem-solving statements. This will greatly help the students in learning how to solve exam questions and display their responses with correct logic. As a result, kids must make the most of the chapter's content by practicing the examples and solving all the questions.

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## Important Questions Class 8 Maths Chapter 9

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## Important Questions Class 8 Mathematics Chapter 9 – Algebraic Expressions and Identities

Algebra is one of the wide areas of Mathematics that studies mathematical expressions involving variables, constants, and mathematical operations like addition, subtraction, multiplication, and division. Algebra is the basis for higher mathematics studies in many fields, including science, engineering, medicine, and economics.

Quick Links

Class 8 Mathematics Chapter 9 – Algebraic expressions and identities is the continuation of the earlier classes and chapters, where students learned about basic algebraic expressions like x + 3, 2y – 5, 3x², 4xy + 7 etc. Many more complex expressions can be formed using variables and constants. The topics covered in Chapter 9 are discussed in the points below.

- Expressions: The problem statement is formed by combining variables, constants and mathematical operators.
- Variables and constants: Consider the example 2x – 3. This expression is formed from the variable X and constants 2 and 3. With different values of the variable x, the result of the expression 2x – 3 changes.
- Terms, factors, and coefficients: Terms in the expression are the product of its factors. The expression 2x – 3 is formed by two terms, 2x and 3. The term 2x is the product of its factors 2 and x. The term 3 comprises just one factor, i.e., 3.

A coefficient is the numerical factor of the term. In the expression 5xy + 3z, the coefficient of the term 5xy is 5, and the coefficient of the term 3z is 3.

- Types of expressions: The three main types of algebraic expressions are,

Monomials: Expression formed with only one term. For example, 4x², -5z, 9xz², etc.

Binomials: Expression formed with two terms. For example 5x – 2, 6y – 9x², a + b, etc

Polynomials: A polynomial is an expression containing one or more terms with a non-zero coefficient and variables having non-negative integers as exponents. For example 2xy, 6x³ + 4x³ + 3x – 1, etc.

- Like and unlike terms: Like terms have the same variables and exponents. However, the coefficients need not be the same. Unlike terms are two or more terms that do not have the same variables or exponents. The order of the variables does not matter unless there is power. To understand better, check the following examples of both types.

Like terms: 4b + 10b, 2x² + 4x², 15w – 13.4w, etc.

Unlike terms: 2x + 9y, 10a² – 9b, x³ – x², etc.

- Subtraction and addition of algebraic expressions: While adding or subtracting polynomials, first add or subtract the like terms; and then handle the unlike terms.
- Multiplication of algebraic expressions: There are three ways to multiply expressions:

(1) A monomial multiplied by a monomial,

(2) A polynomial multiplied by a monomial and

(3) A polynomial multiplied by a polynomial

- Identity: An algebra identity is an equality, which means that the left-hand side of the equation is equal to the right-hand side for all values of the variables. Identities are very useful for solving equations in simple and easy steps. The four standard identities are,
- (a + b)² = a² + 2ab + b²
- (a – b)² = a² – 2ab + b²
- (a + b) (a – b) = a² – b²
- (x + a) (x + b) = x² + (a + b)x + ab

Algebra is an overwhelming and difficult chapter for many students, requiring critical thinking and creative problem-solving skills. Extramarks is one of the most trusted online learning platforms to help students understand the concepts clearly and prepare well for the examinations. Our highly experienced and expert Mathematics teachers have prepared CBSE Mathematics preparation materials for Class 8 students. These study resources include NCERT solutions, solved question papers, sample papers, CBSE revision notes, and more, per the CBSE syllabus and NCERT guidelines.

## Get Access to CBSE Class 8 Maths Important Questions 2022-23 with Chapter-Wise Solutions

You can also find CBSE Class 8 Maths Chapter-by-Chapter Important Questions here:

## Important Questions Class 8 Mathematics Chapter 9 – With Solutions

A complete list of Important Questions Class 8 Mathematics Chapter 9 provided on the Extramarks website consists of various questions for students to analyse their weak areas and practise until they thoroughly understand all the concepts and reduce mistakes. Chapter 9 Class 8 Mathematics Important Questions includes short answer questions, long answer questions, and MCQs backed with detailed and easy-to-understand solutions. These important questions are collated from the NCERT textbook and other genuine sources. By practising from solved question papers, sample papers, and the Mathematics Class 8 Chapter 9 Important Questions, students get familiar with the exam pattern and improve their time-management skills. Regular practise improves the calculating speed, which is helpful in the examination.

Register on the Extramarks website to refer to and practise the Important Questions Class 8 Mathematics Chapter 9 and other chapter-wise questions.

Below is a sample list of questions and their solutions from our Class 8 Mathematics Chapter 9 Important Questions:

Question 1. Find the terms and their coefficients for each of the following expressions.

(i) 5xyz² – 3zy

(ii) 1 + x+ x²

(iii) 4x²y² – 4x²y²z² + z²

(iv) 3- pq + qr – rp

(v) x/2 + y/2 – xy

(vi) 0.3a – 0.6ab + 0.5b

Answer 1. The terms and coefficients are given below,

Question 2. Classify the following polynomials as monomials, binomials, and trinomials. Also, state the

polynomials do not fall in any of these three categories?

x + y, 1000, x + x² + x³, 7 + y + 5x, 2y – 3y², 2y – 3y² + 4y³, 5x – 4y + 3xy, 4z – 15z², ab + bc + cd + da, pqr,

p²q + pq², 2p + 2q,

Answer 2. The classified terms are given below,

Monomials- 1000, pqr

Binomials- x + y, 2y – 3y², 4z – 15z², p²q + pq², 2p + 2q

Trinomials- x + x²+ x³, 7 + y + 5x, 2y – 3y² + 4y³, 5x – 4y + 3xy

Polynomials that do not fall in any of these categories are x + y, x + x²+ x³, ab + bc + cd + da

Question 3. Add the following.

(i) ab – bc, bc – ca, ca – ab

(ii) a – b + ab, b – c + bc, c – a + ac

(iii) 2p²q² – 3pq + 4, 5 + 7pq – 3p²q²

(iv) l² + m², m² + n², n² + l², 2lm + 2mn + 2nl

Answer 3. (i) (ab – bc) + (bc – ca) + (ca – ab)

= ab – bc + bc – ca + ca – ab

= ab-ab-bc+bc-ca+ca

= 0

(ii) (a – b + ab) + (b – c + bc) + (c – a + ac)

= a – b + ab + b – c + bc + c -a +ac

= a – a -b + b + ab – c + c + bc + ac

= ab + bc + ac

(iii) ( 2p²q² – 3pq + 4) + ( 5 + 7pq – 3p²q²)

= 2p²q² – 3pq + 4 + 5 + 7pq – 3p²q²

= 2p²q² – 3p²q² + 7pq – 3pq + 4 + 5

= -1p²q² + 4pq + 9

= 4pq + 9 – p²q²

(iv) ( l² + m²) + (m² + n²) + (n² + l²) + (2lm + 2mn + 2nl)

= l² + m² + m² + n² + n² + l² + 2lm + 2mn + 2nl

= l² + l² + m² + m² + n² + n² + 2lm + 2mn + 2nl

= 2l² + 2m² + 2n² + 2lm + 2mn + 2nl

= 2( l² + m² + n² + lm + mn + nl)

Question 4. Subtract the following.

(i) 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

(ii 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz

(iii) 4p²q – 3pq + 5pq²– 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2+ 5p²q

(i) ( 12a – 9ab + 5b – 3 ) – ( 4a – 7ab + 3b + 12 )

= 12a – 9ab + 5b – 3 – 4a + 7ab – 3b – 12

= 12a – 4a – 9ab + 7ab + 5b –3b – 3 – 12

= 8a – 2ab + 2b – 15

(ii) ( 5xy – 2yz – 2zx + 10xyz ) – ( 3xy + 5yz – 7zx )

= 5xy – 2yz – 2zx + 10xyz – 3xy – 5yz + 7zx

= 5xy – 3xy – 2yz – 5yz – 2zx + 7zx + 10xyz

= 2xy – 7yz + 5zx + 10xyz

(iii) ( 18 – 3p – 11q + 5pq – 2pq2+ 5p²q ) – ( 4p²q – 3pq + 5pq²– 8p + 7q – 10 )

= 18 – 3p – 11q + 5pq – 2pq2 + 5p²q – 4p²q + 3pq – 5pq² + 8p – 7q + 10

= 18 + 10 – 3p + 8p – 11q – 7q + 5pq + 3pq – 2pq2 – 5pq² + 5p²q – 4p²q

= 28 + 5p – 18q + 8pq – 7pq² + p²q

Question 5. Multiply the following.

(i) – 7pq²r³, – 13p³q²r

(ii) 3x²y²z², 17xyz

(iii) 15xy², 17yz²

(iv) –5a²bc, 11ab, 13abc²

(v) (pq – 2r), (pq – 2r)

(vi) (3/2p² + 2/3q²), (2p² –3q²)

Answer 5. (i) ( – 7pq²r³ ) x ( – 13p³q²r )

= 91P 4 q 4 r 4

(ii) ( 3x²y²z² ) x ( 17xyz )

= 51x³y³z³

(iii) ( 15xy² ) x ( 17yz² )

= 255xy³z²

(iv) ( –5a²bc ) x ( 11ab ) x ( 13abc² )

= -715a⁴b³c³

(v) ( pq – 2r ) x ( pq – 2r )

= pq ( pq – 2r) – 2r( pq – 2r )

= p²q² – 2pqr – 2rpq + 4r²

= p²q² – 4pqr + 4r²

(vi) ( 3 p² + 2 q² ) x ( 2p² – 3q² )

2 3

= 3p² x 2p² – 3 p² x 3q² + 2q² x 2p² – 2q² x 3q²

2 2 3 3

= 6P 4 – 9p²q² + 4q²p² – 6q 4

2 2 3 3

= 3P 4 – 9p²q² + 4q²p² – 2q 4

2 3

Question 6. Which term is the like term similar to 24a²bc?

(a) 13 × 8a × 2b × c × a

(b) 8 × 3 × a × b × c

(c) 3 × 8 × a × b × c × c

(d) 3 × 8 × a × b × b × c

Answer 6. Option (a)

Explanation: To find out the similar term as 24a²bc, let us find the product of each of the equations,

- 13 × 8a × 2b × c × a = 208a²bc
- 8 × 3 × a × b × c = 24abc
- 3 × 8 × a × b × c × c = 24abc²
- 3 × 8 × a × b × b × c = 24ab²c

Hence, we can get that option (a) is correct.

Question 7. Which of the following is an identity?

(a) (p + q)² = p² + q²

(b) p² – q² = (p – q)²

(c) p² – q² = p² + 2pq – q²

(d) (p + q)² = p² + 2pq + q²

Answer 7. Option (d)

Explanation: The equation (p + q)² = p² + 2pq + q² follows the first standard algebraic identity

( a + b )² = a² + 2ab + b². The rest of the other options do not follow any of the standard identities. Hence option (d) is correct.

Question 8. Fill in the blanks.

(a) (x + a) (x + b) = x² + (a + b)x + ________.

(b) The product of two terms with like signs is a ________ term.

(c) The product of two terms with unlike signs is a ________ term.

(d) (a – b) _________ = a² – 2ab + b²

(e) a² – b² = (a + b ) __________.

(f) (a – b)² + ____________ = a² – b²

(g) (a + b)² – 2ab = ___________ + ____________

(h) The product of two polynomials is a ________

(i) The coefficient in – 37abc is __________.

(j) Number of terms in the expression a2 + bc × d is ________

As per the standard identity 4, (x + a) (x + b) = x² + (a + b)x + ab

(b) Positive

(c) Negative

(d) ^2 or ( a – b)²

As per standard identity 2, (a – b)² = a² – 2ab + b²

(e) (a – b)

As per standard identity 3, (a + b ) ( a – b ) = a² – b²

(f) 2ab – 2b²

Let us solve the equation with x in the blank space. As per identity 2, (a – b)² = a² – 2ab + b².

Hence, a² – 2ab + b² + x = a² – b²

x = a² – b² – a² + 2ab – b²

x = 2ab – 2b²

(g) a² + b²

Using Identity 1 ( a + b )² = a² + 2ab + b²,

(a + b)² – 2ab = a² + 2ab + b² – 2ab = a² + b²

(h) Polynomial

Question 9. Solve the below using correct identities.

(b) 181² – 19²

(c) 497 × 505

(d) 2.07 × 1.93

= (50 – 2)²

As (a – b)² = a² – 2ab + b² , hence

(50 – 2)² = (50)² – 2 × 50 × 2 + (2)²

= 2500 – 200 + 4

= 2300 + 4

= 2304

(b) As a² – b² = (a – b) (a + b)

181² – 19² = (181 – 19) (181 + 19)

= 162 × 200

= 32400

(c) By using the identity (x + a) (x + b) = x2 + (a + b) x + ab

497 x 505 = ( 500 – 3 ) (500 + 5 )

= 500² + (–3 + 5) × 500 + (–3) (5)

= 250000 + 1000 – 15

= 250985

(d) As (a + b) (a – b) = a² – b²

2.07 × 1.93 = (2 + 0.07) (2 – 0.07)

= 2² – 0.07²

= 3.9951

Question 10. The length of a rectangular box is ( x + 9y) and the area is x² + 12xy + 27y². Find the breadth.

Answer 10. Area of a rectangle = length x breadth, hence breadth = area / length.

breadth = x² + 12xy + 27y²

( x + 9y )

= x² + 9xy + 3xy + 27y²

( x + 9y )

= x ( x + 9y ) + 3y (x + 9y)

( x + 9y )

breadth = x + 3y

Question 11. With the help of identity (x + a) (x + b) = x² + (a + b) x + ab, find the following products.

(a) (x + 3) (x + 7)

(b) (4x + 5) (4x + 1)

(c) (4x – 5) (4x – 1)

(d) (4x + 5) (4x – 1)

(e) (2x + 5y) (2x + 3y)

(f) (2a²+ 9) (2a²+ 5)

(g) (xyz – 4) (xyz – 2)

Answer 11.

- ( x + 3 ) ( x + 7 )

= x² + ( 3 + 7 ) x + ( 3 x 7 )

= x² + 10x + 21

- ( 4x + 5 ) ( 4x + 1 )

= 16x² + ( 5 + 1 ) 4x + ( 5 x 1 )

= 16x² + 24x + 5

(c) (4x – 5) (4x – 1)

= 16x² + ( – 5 – 1 ) 4x + ( -5 x -1 )

= 16x² – 24x + 5

(d) (4x + 5) (4x – 1)

= 16x² + ( 5 + ( – 1 ) )4x + ( 5 x ( – 1 ) )

= 16x² + 16x – 5

(e) (2x + 5y) (2x + 3y)

= 4x² + ( 5y + 3y )2x + ( 5y x 3y )

= 4x² + 16xy + 15y²

(f) (2a²+ 9) (2a²+ 5)

= 4a 4 + ( 9 + 5 )2a² + ( 9 x 5 )

= 4a 4 + 28a² + 45

(g) (xyz – 4) (xyz – 2)

= x²y²z² + ( -4 – 2)xyz + ( – 4 x – 2)

= x²y²z² – 6xyz + 8

Question 12. The exponents of the variables in a polynomial are always

(a) Integers

(b) Positive integers

(c) Non-negative integers

(d) Non-positive integers

Answer 12. (c) Non-negative integers

Explanation: A polynomial will have a non-zero coefficient and variables having non-negative integers as exponents. For example : a + b + r + q, 3ab, 5xyz – 10, 2a + 3b + 7z, etc.

Question 13. The product of a binomial and monomial is a

(a) Monomial

(b) Binomial

(c) Trinomial

(d) None of these

Answer 13. (b) Binomial

Explanation: This can be demonstrated through an example below,

x ( y + z ) = xy + xz

This expression contains two terms, x is a monomial and ( y + z ) is a binomial.

The product of multiplying these terms results in a binomial product xy + xz.

Question 14. Using identities, find products for the below.

(f) 297 × 303

(g) 78 × 82

(i) 10.5 × 9.5

Answer 14.

- 71² = ( 70 + 1 )² Identity applied ( a + b )² = a² + 2ab + b²

= 70² + 2 ( 70 x 1 ) + 1²

= 4900 + 140 + 1

- 99² = ( 100 – 1 )² Identity applied ( a – b )² = a² – 2ab + b²

= 100² – 2 ( 100 x 1 ) + 1²

= 10000 – 200 + 1

= 9801

(c) 102² = ( 100 + 2 )² Identity applied ( a + b )² = a² + 2ab + b²

= 100² + 2 ( 100 x 2 ) + 2²

= 10000 + 400 + 4

= 10404

(d) 998² = ( 1000 – 2 )² Identity applied ( a – b )² = a² – 2ab + b²

= 1000² – 2 ( 1000 x 2 ) + 2²

= 1000000 – 4000 + 4

= 996004

(e) 5.2² = ( 5 + 0.2 )² Identity applied ( a + b )² = a² + 2ab + b²

= 5² + 2 ( 5 x 0.2 ) + 0.2²

= 25 + 2 + 0.04

= 27.04

(f) 297 × 303 = ( 300 – 3 ) ( 300 + 3 ) Identity applied ( a + b ) ( a – b ) = a² – b²

= 300² – 3²

= 90000 – 9

= 89991

(g) 78 × 82 = ( 80 – 2 ) ( 80 + 2 ) Identity applied ( a + b ) ( a – b ) = a² – b²

= 80² – 2²

= 6400 – 4

= 6396

(h) 8.9² = ( 9.0 – 0.1 )² Identity applied ( a – b )² = a² – 2ab + b²

= 9.0² – 2 ( 9.0 x 0.1 ) + 0.1²

= 81 – 1.8 + 0.01

= 79.21

(i) 10.5 x 9.5 = ( 10 + 0.5 ) ( 10 – 0.5 ) Identity applied ( a + b ) ( a – b ) = a² – b²

= 10² – 0.5²

= 100 – 0.25

= 99.75

Question 15. The Coefficient of y in the term −y is

3

3

3

Answer 15. (c) – 1

3

Explanation: Coefficient is defined as the numerical factor of a term.

Hence, the numerical factor/ coefficient of the term −y is −1

3 3

Question 16.Obtain the volume of rectangular boxes with the following length, breadth and height

respectively.

(a) 5a, 3a², 7a 4

(b) 2p, 4q, 8r

(c) xy, 2x²y, 2xy²

(d) a, 2b, 3c

Answer 16. The volume of a rectangular box is the product of its length,

breadth and height, i.e Volume = length x breadth x height.

Volumes are calculated as below,

(a) length = 5a, breadth = 3a², height = 7a 4

Volume = 5a x 3a² x 7a 4

(b) length = 2p, breadth = 4q, height = 8r

Volume = 2p x 4q x 8r

(c) length = xy, breadth = 2x²y, height = 2xy²

Volume = xy x 2x²y x 2xy²

(d) length = a, breadth = 2b, height = 3c

Volume = a x 2b x 3c

Question 17. State whether the following are True (T) or False (F):

(a) (a + b)² = a² + b²

(b) (a – b)² = a² – b²

(c) (a + b) (a – b) = a² – b²

(d) The product of two negative terms is a negative term.

(e) The product of one negative and one positive term is a negative term.

(f) The coefficient of the term – 6x²y² is – 6.

(g) p²q + q²r + r²q is a binomial

(h) An equation is true for all values of its variables.

Answer 17.

(a) False. ( a + b )² = a² + 2ab + b²

(b) False. ( a – b )² = a² – 2ab + b²

(d) False. The product of two negative terms is positive.

(g) False. The equation p²q + q²r + r²q consists of three terms; hence it is a trinomial.

(h) False. An equation is not true for all values of its variables. For example : 4x + 2 = 10 is true, only for x = 2.

Question 18. Show that LHS = RHS for the below equations.

(a) ( 3x + 7 )² – 84x = ( 3x – 7 )²

(b) ( 9p – 5q )² + 180pq = ( 9p + 5q )²

(c) ( 4pq + 3q )² – ( 4pq – 3q )² = 48pq²

(d) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

Answer 18.

(a) LHS = ( 3x + 7 )² – 84x

= (3x)² + 2 ( 3x x 7 ) + 7² – 84x

= 9x² + 42x + 49 – 84x

= 9x² – 42x + 49

RHS = ( 3x – 7 )²

= (3x)² – 2 ( 3x x 7 ) + 7²

Hence LHS = RHS

(b) LHS = ( 9p – 5q )² + 180pq

= (9p)² – 2 ( 9p x 5q ) + (5q)² + 180pq

= 81p² + 90pq + 25q²

RHS = ( 9p + 5q )²

= (9p)² + 2 ( 9p x 5q ) + (5q)²

= 81p² + 90pq + 25q²

(c) LHS = ( 4pq + 3q )² – ( 4pq – 3q )²

= (4pq)² + 2 ( 4pq x 3q ) + (3q)² – ( (4pq)² – 2 ( 4pq x 3q ) + (3q)²)

= 16p²q² + 24pq² + 9q² – 16p²q² + 24pq² – 9q²

= 48pq²

RHS = 48pq²

Hence LHS = RHS

(d) LHS = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)

= a² + ab – ba – b² + b² + bc – cb – c² + c² + ca – ac – a²

= 0

RHS = 0

Question 19. Expand the following, using suitable identities.

(a) ( xy + yz )²

(b) ( x²y – xy² )²

(c) ( 7x + 5 )²

(d) ( 0.9p – 0.5q )²

(e) ( x² + y² ) ( x² – y² )

Answer 19.

(a) (xy + yz)²

= (xy)² + 2 (xy x yz ) + (yz)²

= x²y² + 2xy²z + y²z²

(b) (x²y – xy²)²

= (x²y)² – 2 (x²y x xy²) + (xy²)²

= x 4 y² – 2x³y³ + x²y 4

(c) (7x + 5)²

= (7x)² + 2 (7x x 5) + (5)²

= 49x² + 70x + 25

(d) (0.9p – 0.5q)²

= (0.9p)² – 2 ( 0.9p x 0.5q) + (0.5q)²

= 0.81p² – 0.9pq + 0.25q²

= x 4 – x²y² + y²x² – y 4

= x 4 – y 4

Question 20. Select the correct option of volume of a rectangular box with length = 2ab, breadth = 3ac and height = 2ac

(a) 12a³bc²

(d) 2ab +3ac + 2ac

Answer 20. Option (a)

Explanation: The formula for calculating the volume of a rectangular box is

Volume = length x breadth x height

With the length of the input = 2ab, breadth = 3ac and height = 2ac

Volume = 2ab x 3ac x 2ac

= 12a³bc²

## Benefits of Solving Important Questions Class 8 Mathematics Chapter 9

Practising the questions from Important Questions Class 8 Mathematics Chapter 9 available on the Extramarks website will benefit the students in grasping the chapter concepts like variables, terms, and standard identities easily and solving the equations with better understanding and avoiding mistakes. Solving the important questions of the chapter rigorously will help you score well in the examination.

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- The study and practice materials are based on the latest versions of the CBSE syllabus and NCERT guidelines. Students can rely upon these materials to get well versed in the examination’s basic question/answer format.
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- By thoroughly practising these tricky questions from the Important Questions Class 8 Mathematics Chapter 9, students can analyse their weak areas of Chapter 9 Algebra and overcome them before facing their final examinations.
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Q.1 Using identity, find the value of (4.7) 2 .

Marks: 2 Ans

Use the identity as (a b) 2 = a 2 2ab + b 2 .

4.7 2 = 5 0.3 2 = 5 2 2 — 5 — 0.3 + 0.3 2 = 25 3.0 + 0.09 = 22.09

Hence, the required value is 22.09.

Q.2 Simplify: (a + 2) (c 4) + (a 2) (c + 2) + 2 (ac + 17)

(a + 2) (c 4) + (a 2) (c + 2) + 2 (ac + 17) = ac 4a + 2c 8 + ac + 2a 2c 4 + 2ac + 34 = 4ac 2a + 22

Q.3 The value of 1001 × 1004 is

Marks: 1 Ans [ 2049078 ]

Q.4 If the length of a rectangle is 2x 2 + 3 units and its breadth is 2x 2 7 units. What is the area and perimeter of the rectangle

Marks: 4 Ans

(x + a) (x + b) = x 2 + x (a + b) + ab Area = length × breadth = (2x 2 + 3) × (2x 2 7) = (2x 2 ) 2 + 2x 2 (3 7) + 3(7) = 4x 4 8x 2 21 square units Perimeter = 2(length + breadth) = 2(2x 2 + 3 + 2x 2 7) = 2(4x 2 4) = 8x 2 8 units

If x 2 + 1 x 2 = 4 , what is the value of x + 1 x

Given: x 2 + 1 x 2 = 4 using identities: (a + b) 2 = a 2 + b 2 + 2ab x + 1 x 2 = x 2 + 1 x 2 + 2 x × 1 x = 4 + 2 = 6

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Cbse class 8 maths important questions, chapter 1 - rational numbers.

## Chapter 2 - Linear Equations in One Variable

Chapter 3 - understanding quadrilaterals, chapter 4 - practical geometry, chapter 5 - data handling, chapter 6 - squares and square roots, chapter 7 - cubes and cube roots, chapter 8 - comparing quantities, chapter 10 - visualising solid shapes, chapter 11 - mensuration, chapter 12 - exponents and powers, chapter 13 - direct and inverse proportions, chapter 14 - factorisation, chapter 15 - introduction to graphs, chapter 16 - playing with numbers, faqs (frequently asked questions), 1. how can i score good marks in class 8 mathematics.

One of the most effective ways to understand and remember the concepts is by practising a variety of questions at the end of the chapter. The Extramarks team understands the importance of practising the questions while preparing for the examinations and provides a full list of Important Questions Class 8 Mathematics Chapter 9. Students can practise these advanced levels of questions on the Extramarks website to become familiar with different types of questions and perform well in their exams.

## 2. What do the Important Questions Class 8 Mathematics Chapter 9 include?

The Important Questions Class 8 Mathematics Chapter 9 includes a variety of easy as well as difficult questions from different sources like the NCERT textbook, NCERT solutions, NCERT Exemplars, and other standard books for students to practise and excel in their school and competitive examinations. The questions come along with detailed step-by-step solutions, which are created by Extramarks Mathematics experts who completely feel the pulse of the students while preparing such a questionnaire. Each and every topic has been taken care of to avoid unnecessary stress and anxiety students might face while learning. It gives the students enough practice to boost their confidence level. The more you practise, the better you will get. That’s the key to success.

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## NCERT Solutions for Class 8 Maths Chapter 9 - Algebraic Expressions And Identities

- NCERT Solutions
- Chapter 9 Algebraic Expressions And Identities

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities - Free PDF

Vedantu’s NCERT Solutions Class 8 Maths is like an all-time available solution to the problems of students whether it is a Class test or the final exam. These NCERT solutions are the best solutions for the questions given in the NCERT books and will help the students to get the best marks in exams. To get the best and better solution, you may always rely upon our NCERT Solution Class 8 Chapter 9 as they are prepared in a very easy to understand language by our subject experts. All the Mathematical steps of solutions to different maths problems are explained in a language that is easily comprehensible by the students. Science Students who are looking for NCERT Solutions for Class 8 Science will also find the Solutions curated by our Master Teachers really Helpful. You can also download NCERT Solutions Class 8 Maths to help you to revise complete syllabus ans score more marks in your examinations. Learning has never been so simple!

## Topics Covered Under NCERT Class 8 Math Chapter 9 Algebraic Expressions and Identities

Chapter 9 Algebraic Expressions and Identities is an important topic for Class 8 as well as for higher mathematics. Algebra is an interesting concept of mathematics, and the NCERT Class 8 Math Chapter 9 Algebraic Expressions and Identities covers an introduction to this unit in detail.

With several identities, expressions, examples, and exercise questions, Chapter 9 of NCERT Class 8 Math is a lengthy one. So, students can maintain their motivation by knowing beforehand what topics would be covered in each exercise of the chapter.

The important concepts covered under Math Class 8 Algebraic Expressions and Identities can be seen in the table below.

## Access NCERT Solutions for Class 8 Mathematics Chapter 9 – Algebraic Expressions and Identities

Exercise – 9.1

1. Identify the terms, their coefficient for each of the following expressions.

(i) $5{\text{xy}}{{\text{z}}^2} - 3{\text{zy}}$

Ans:

(ii) $1 + {\text{x}} + {{\text{x}}^2}$

(iii) ${\text{4}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}{\text{ - 4}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}{{\text{z}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}$

(iv) ${\text{3 - pq + qr - rp}}$

Ans:

(v) $\dfrac{{\text{x}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{y}}}{{\text{2}}}{\text{ - xy}}$

Ans:

(vi) ${\text{0}}{\text{.3a - 0}}{\text{.6ab + 0}}{\text{.5b}}$

2. Classify the following polynomials as monomials, binomials, trinomials. In which polynomials do not fit in any of these three categories?

${x + y}$, 1000, ${x + {{x}^{2}} + {{x}^{3}} + {{x}^{4}}}$, ${7 + y + 5x}$, ${2y - 3{y}^{2}}$, ${2y - 3{y}^{2} + 4{y}^{3}}$, ${5x - 4y + 3xy}$, ${4z -15{z}^{2}}$, ${ab + bc + cd + da}$, pqr, ${{p}^{2}q + p{q}^{2}}$, ${2p + 2q}$

Ans: The given expressions are classified as

Monomials: ${\text{1000,}}$${\text{pqr}}$

Binomials: ${\text{x + y,2y - 3}}{{\text{y}}^{\text{2}}}{\text{,4z - 15}}{{\text{z}}^{\text{2}}}{\text{,}}{{\text{p}}^{\text{2}}}{\text{q + p}}{{\text{q}}^{\text{2}}}{\text{,2p + 2q}}$

Trinomials: ${\text{7 + y + 5x,2y - 3}}{{\text{y}}^{\text{2}}}{\text{ + 4}}{{\text{y}}^{\text{3}}}{\text{,5x - 4y + 3xy}}$

Polynomials that do not fit in any categories are

${\text{x + }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{x}}^{\text{3}}}{\text{ + }}{{\text{x}}^{\text{4}}}{\text{,ab + bc + cd + da}}$

3. Add the following:

(i) ${\text{ab - bc,bc - ca,ca - ab}}$

${\text{ 12a - 9ab + 5b - 3}} $

Therefore, the sum of the given expressions is o.

(ii) ${\text{a - b + ab,b - c + bc,c - a + ac}}$

$ {\text{ }}a - b + ab $

$ {\text{ }} + b{\text{ }} - c + bc $

$ {\text{ }} + \quad - a{\text{ }} + c{\text{ + ac}} $

$ \overline {{\text{ ab + bc + ac}}} $

Thus the sum of given expressions is ${\text{ab + bc + ac}}$

(iii) ${\text{2}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{ - 3pq + 4,5 + 7pq - 3}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}$

$ {\text{ 2}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{ - 3pq + 4}} $

$ {\text{ + - 3}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{ + 7pq + 5}}$

$ \overline {{\text{ - }}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{ + 4pq + 9}}} $

Therefore, the sum of given expressions is ${\text{ - }}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{ + 4pq + 9}}$

(iv) ${{\text{l}}^{\text{2}}}{\text{ + }}{{\text{m}}^{\text{2}}}{\text{,}}{{\text{m}}^{\text{2}}}{\text{ + }}{{\text{n}}^{\text{2}}}{\text{,}}{{\text{n}}^{\text{2}}}{\text{ + }}{{\text{l}}^{\text{2}}}{\text{,2lm + 2mn + 2nl}}$

$ {\text{ }}{{\text{l}}^{\text{2}}}{\text{ + }}{{\text{m}}^{\text{2}}} $

$ {\text{ + }}{{\text{m}}^{\text{2}}}{\text{ + }}{{\text{n}}^{\text{2}}} $

$ {\text{ + }}{{\text{l}}^{\text{2}}}{\text{ + }}{{\text{n}}^{\text{2}}} $

$ {\text{ + 2lm + 2mn + 2nl}} $

$ \overline {{\text{ 2}}{{\text{l}}^{^{\text{2}}}}{\text{ + 2}}{{\text{m}}^{\text{2}}}{\text{ + 2}}{{\text{n}}^{\text{2}}}{\text{ + 2lm + 2mn + 2nl}}} $

Therefore, the sum of the given expressions is ${\text{2}}{{\text{l}}^{^{\text{2}}}}{\text{ + 2}}{{\text{m}}^{\text{2}}}{\text{ + 2}}{{\text{n}}^{\text{2}}}{\text{ + 2lm + 2mn + 2nl}}$

4. Solve the following:

(i) Subtract ${\text{4a - 7ab + 3b + 12}}$ from ${\text{12a - 9ab + 5b - 3}}$

$ {12a - 9ab + 5b - 3} $

$ {4a - 7ab + 3b + 12} $

$ {( - )\quad ( + )\quad ( - )( - )} $

$ {\overline {8a - 2ab + 2b - 15} } $

(ii) Subtract ${\text{3xy + 5yz - 7zx}}$ from ${\text{5xy - 2yz - 2zx + 10xyz}}$

$ {\text{5xy - 2yz - 2zx + 10xyz}} $

$ {\text{3xy + 5yz - 7zx}} $

$ {\text{( - )( - )}}\quad {\text{( + )}} $

$ \overline {{\text{2xy - 7yz + 5zx + 10xyz}}} $

(iii) Subtract ${\text{4p 2q - 3pq + 5pq2 - 8p + 7q - 10}}$from ${\text{18 - 3p - 11q + 5pq - 2pq2 + 5p 2q}}$

$ {\text{18 - 3p - 11q + 5pq - 2p}}{{\text{q}}^{\text{2}}}{\text{ + 5}}{{\text{p}}^{\text{2}}}{\text{q}} $

$ {\text{ - 10 - 8p + 7q - 3pq + 5p}}{{\text{q}}^{\text{2}}}{\text{ + 4}}{{\text{p}}^{\text{2}}}{\text{q}} $

$ \dfrac{{{\text{( + )( + )( - )( + )( - )}}\quad {\text{( - )}}}}{{{\text{28 + 5p - 18q + 8pq - 7p}}{{\text{q}}^{\text{2}}}{\text{ + }}{{\text{p}}^{\text{2}}}{\text{q}}}} $

Exercise – 9.2

1. Find the product of the following pairs of monomials.

(i) ${\text{4,7p}}$

Ans: ${{4 \times 7p = 4 \times 7 \times p = 28p}}$

(ii) $\dfrac{{{\text{first monomial}} \to }}{{{\text{second monomial}} \downarrow }}$

Ans: ${{ - 4p \times 7p = - 4 \times p \times 7 \times p = }}\left( {{{ - 4 \times 7}}} \right){{ \times }}\left( {{{p \times p}}} \right){\text{ = - 28 }}{{\text{p}}^2}$

(iii) ${\text{ - 4p,7pq}}$

Ans: ${{ - 4p \times 7pq = - 4 \times p \times 7 \times p \times q = }}\left( {{{ - 4 \times 7}}} \right){{ \times }}\left( {{{p \times p \times q}}} \right){\text{ = - 28}}{{\text{p}}^2}{\text{q }}$

(iv) ${\text{4}}{{\text{p}}^{\text{3}}}{\text{ , - 3p }}$

Ans: ${\text{ 4}}{{\text{p}}^{\text{3}}}{{ \times - 3p = 4 \times }}\left( {{\text{ - 3}}} \right){{ \times p \times p \times p \times p = - 12 }}{{\text{p}}^{\text{4}}}$

(v) ${\text{4p, 0}}$

Ans: ${{4p \times 0 = 4 \times p \times 0 = 0 }}$

2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

$\left( {{\text{p, q}}} \right){\text{; }}\left( {{\text{10m, 5n}}} \right){\text{; }}\left( {{\text{20}}{{\text{x}}^{\text{2}}}{\text{ , 5}}{{\text{y}}^{\text{2}}}{\text{ }}} \right){\text{; }}\left( {{\text{4x, 3}}{{\text{x}}^{\text{2}}}{\text{ }}} \right){\text{; }}\left( {{\text{3mn, 4np}}} \right){\text{ }}$

Ans: We know that,

Area of rectangle = length x breadth

Area of 1 st rectangle = p x q = pq

Area of 2 nd rectangle = ${{10m \times 5n = 10 \times 5 \times m \times n = 50mn}}$

Area of 3 rd rectangle = ${\text{20}}{{\text{x}}^{\text{2}}}{{ \times 5}}{{\text{y}}^{\text{2}}}{{ = 20 \times 5 \times }}{{\text{x}}^{\text{2}}}{{ \times }}{{\text{y}}^{\text{2}}}{\text{ = 100}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}$

Area of 4 th rectangle = ${{4x }} \times {\text{ 3}}{{\text{x}}^{\text{2}}}{{ = 4 \times 3}} \times {{x}} \times {{\text{x}}^2}{\text{ = 12}}{{\text{x}}^3}$

Area of 5 th rectangle ${{ = 3mn \times 4np = 3 \times 4 \times m \times n \times n \times p = 12m}}{{\text{n}}^{\text{2}}}{\text{p}}$

3. Complete the table of products.

The table can be completed as follows.

4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(i) ${\text{5a,3}}{{\text{a}}^{\text{2}}}{\text{,7}}{{\text{a}}^{\text{4}}}$

Ans: We know that

Volume= length x breadth x height

Volume = ${{5a \times 3}}{{\text{a}}^{\text{2}}}{{ \times 7}}{{\text{a}}^{\text{4}}}{\text{ = 105}}{{\text{a}}^{\text{7}}}$

(ii) ${\text{2p,4q,8r}}$

Volume = length x breadth x height

Volume = ${{2p \times 4q \times 8r = 64pqr}}$

(iii) ${\text{xy,2}}{{\text{x}}^{\text{2}}}{\text{y,2x}}{{\text{y}}^{\text{2}}}$

Volume = ${{xy \times 2}}{{\text{x}}^{\text{2}}}{{y \times 2x}}{{\text{y}}^{\text{2}}}{\text{ = 4}}{{\text{x}}^{\text{4}}}{{\text{y}}^{\text{4}}}$

(iv) ${\text{a,2b,3c}}$

Volume = ${{a}} \times {\text{2b}} \times {\text{3c = 6abc}}$

5. Obtain the product of

(i) ${\text{xy, yz, zx }}$

Ans: ${{xy \times yz \times zx = }}{{\text{x}}^{\text{2}}}{\text{ }}{{\text{y}}^{\text{2}}}{\text{ }}{{\text{z}}^{\text{2}}}$

(ii) ${\text{a, - }}{{\text{a}}^{\text{2}}}{\text{ , }}{{\text{a}}^{\text{3}}}{\text{ }}$

Ans: ${{a}} \times ({\text{ - }}{{\text{a}}^{{2}}}) \times {\text{ }}{{\text{a}}^{\text{3}}}{\text{ = - }}{{\text{a}}^6}{\text{ }}$

(iii) ${\text{2, 4y, 8}}{{\text{y}}^2}{\text{ , 16}}{{\text{y}}^3}$

Ans: ${{2}} \times {{ 4y}} \times {\text{8}}{{\text{y}}^2} \times {\text{ 16}}{{\text{y}}^3} = 1024{y^6}$

(iv) ${\text{a, 2b, 3c, 6abc}}$

Ans: ${{a \times 2b \times 3c \times 6abc = }}$${\text{36}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}$

(v) ${\text{m, - mn, mnp}}$

Ans: ${{m \times }}\left( {{\text{ - mn}}} \right){{ \times mnp = - }}{{\text{m}}^{\text{3}}}{{\text{n}}^{\text{2}}}$

Exercise – 9.3

(i) Carry out the multiplication of the expressions in each of the following pairs.

(i) ${\text{4p, q + r }}$

Ans: $\left( {{\text{4p}}} \right){{ \times }}\left( {{\text{q + r}}} \right){\text{ = }}\left( {{{4p \times q}}} \right){\text{ + }}\left( {{{4p \times r}}} \right){\text{ = 4pq + 4pr}}$

(ii) ${\text{ab, a - b }}$

Ans: $\left( {{\text{ab}}} \right){{ \times }}\left( {{\text{a - b}}} \right){\text{ = }}\left( {{{ab \times a}}} \right){\text{ + }}\left[ {{{ab \times }}\left( {{\text{ - b}}} \right)} \right]{\text{ = }}{{\text{a}}^{\text{2}}}{\text{b - a}}{{\text{b}}^{\text{2}}}$

(iii) ${\text{a + b, 7}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}$

Ans: $\left( {{\text{a + b}}} \right){{ \times }}\left( {{\text{7a 2 b 2 }}} \right){\text{ = }}\left( {{{a \times 7}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{\text{ }}} \right){\text{ + }}\left( {{{b \times 7}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{\text{ }}} \right){\text{ = 7}}{{\text{a}}^{\text{3}}}{{\text{b}}^{\text{2}}}{\text{ + 7}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{3}}}$

(iv) ${{\text{a}}^{\text{2}}}{\text{ - 9, 4a}}$

Ans: $\left( {{{\text{a}}^2}{\text{ - 9}}} \right){{ \times }}\left( {{\text{4a}}} \right){\text{ = }}\left( {{{\text{a}}^{\text{2}}}{{ \times 4a}}} \right){\text{ + }}\left( {{\text{ - 9}}} \right){{ \times }}\left( {{\text{4a}}} \right){\text{ = 4}}{{\text{a}}^{\text{3}}}{\text{ - 36a}}$

(v) ${\text{pq + qr + rp, 0}}$

Ans: $\left( {{\text{pq + qr + rp}}} \right){{ \times 0 = }}\left( {{{pq \times 0}}} \right){\text{ + }}\left( {{{qr \times 0}}} \right){\text{ + }}\left( {{{rp \times 0}}} \right){\text{ = 0 }}$

Complete the table

Ans: The table can be completed as follows:

3. Find the product :

(i) $\left( {{{\text{a}}^{\text{2}}}} \right){{ \times }}\left( {{\text{2}}{{\text{a}}^{{\text{22}}}}} \right){{ \times }}\left( {{\text{4}}{{\text{a}}^{{\text{26}}}}} \right)$

Ans: $\left( {{{\text{a}}^{\text{2}}}} \right){{ \times }}\left( {{\text{2}}{{\text{a}}^{{\text{22}}}}} \right){{ \times }}\left( {{\text{4}}{{\text{a}}^{{\text{26}}}}} \right){{ = 2 \times 4 \times }}{{\text{a}}^{\text{2}}}{{ \times }}{{\text{a}}^{{\text{22}}}}{{ \times }}{{\text{a}}^{{\text{26}}}}{\text{ = 8}}{{\text{a}}^{{\text{50}}}}$

(ii) $\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{xy}}} \right){{ \times }}\left( {\dfrac{{{\text{ - 9}}}}{{{\text{10}}}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}} \right)$

Ans: $\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{xy}}} \right){{ \times }}\left( {\dfrac{{{\text{ - 9}}}}{{{\text{10}}}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}} \right){\text{ = }}\left( {\dfrac{{\text{2}}}{{\text{3}}}} \right){{ \times }}\left( {\dfrac{{{\text{ - 9}}}}{{{\text{10}}}}} \right){{ \times x \times y \times }}{{\text{x}}^{\text{2}}}{{ \times }}{{\text{y}}^{\text{2}}}{\text{ = }}\dfrac{{{\text{ - 3}}}}{{\text{5}}}{{\text{x}}^{\text{3}}}{{\text{y}}^{\text{3}}}$

(iii) $\left( {{\text{ - }}\dfrac{{{\text{10}}}}{{\text{3}}}{\text{p}}{{\text{q}}^{\text{3}}}} \right){{ \times }}\left( {\dfrac{{\text{6}}}{{\text{5}}}{{\text{p}}^{\text{3}}}{\text{q}}} \right)$

Ans: $\left( {{\text{ - }}\dfrac{{{\text{10}}}}{{\text{3}}}{\text{p}}{{\text{q}}^{\text{3}}}} \right){{ \times }}\left( {\dfrac{{\text{6}}}{{\text{5}}}{{\text{p}}^{\text{3}}}{\text{q}}} \right){\text{ = }}\left( {\dfrac{{{\text{ - 10}}}}{{\text{3}}}} \right){{ \times }}\left( {\dfrac{{\text{6}}}{{\text{5}}}} \right){{ \times p}}{{\text{q}}^{\text{3}}}{{ \times }}{{\text{p}}^{\text{3}}}{\text{q = - 4}}{{\text{p}}^{\text{4}}}{{\text{q}}^{\text{4}}}$

(iv) ${{x \times }}{{\text{x}}^{\text{2}}}{{ \times }}{{\text{x}}^{\text{3}}}{{ \times }}{{\text{x}}^{\text{4}}}$

Ans: ${{x \times }}{{\text{x}}^{\text{2}}}{{ \times }}{{\text{x}}^{\text{3}}}{{ \times }}{{\text{x}}^{\text{4}}}{\text{ = }}{{\text{x}}^{10}}$

4. Solve the following

(i) Simplify ${\text{3x }}\left( {{\text{4x - 5}}} \right){\text{ + 3}}$and find its values for

(a) ${\text{ x = 3}}$

Ans: ${\text{3x }}\left( {{\text{4x - 5}}} \right){\text{ + 3 = 12}}{{\text{x}}^{\text{2}}}{\text{ - 15x + 3 }}$

$ {\text{ For x = 3, 12}}{{\text{x}}^{\text{2}}}{\text{ - 15x + 3 = 12 }}{\left( {\text{3}} \right)^{\text{2}}}{\text{ - 15}}\left( {\text{3}} \right){\text{ + 3 }} $

$ {\text{ = 108 - 45 + 3 }} $

$ {\text{ = 66 }} $

(b) ${\text{x = }}\dfrac{{\text{1}}}{{\text{2}}}$

$ {\text{ For x = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{, 12}}{{\text{x}}^{\text{2}}}{\text{ - 15x + 3 = 12 }}{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)^{\text{2}}}{\text{ - 15}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ + 3 }} $

$ {\text{ = 3 - }}\dfrac{{{\text{15}}}}{{\text{2}}}{\text{ + 3 }} $

$ {\text{ = 6 - }}\dfrac{{{\text{15}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{12 - 15}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{ - 3}}}}{{\text{2}}} $

(ii) ${\text{a }}\left( {{{\text{a}}^{\text{2}}}{\text{ + a + 1}}} \right){\text{ + 5}}$ and find its value for

(a) ${\text{a = 0}}$

Ans: ${\text{For a = 0, }}{{\text{a}}^{\text{3}}}{\text{ + }}{{\text{a}}^{\text{2}}}{\text{ + a + 5 = 0 + 0 + 0 + 5 = 5}}$

(b) ${\text{a = 1}}$

Ans: $ {\text{For a = 1, }}{{\text{a}}^{\text{3}}}{\text{ + }}{{\text{a}}^{\text{2}}}{\text{ + a + 5 = }}{\left( {\text{1}} \right)^{\text{3}}}{\text{ + }}{\left( {\text{1}} \right)^{\text{2}}}{\text{ + 1 + 5}} $

$ {\text{ = 1 + 1 + 1 + 5 = 8 }} $

(c) ${\text{a = - 1}}$

Ans: $ {\text{For a = - 1, }}{{\text{a}}^{\text{3}}}{\text{ + }}{{\text{a}}^{\text{2}}}{\text{ + a + 5 = }}{\left( {{\text{ - 1}}} \right)^{\text{3}}}{\text{ + }}{\left( {{\text{ - 1}}} \right)^{\text{2}}}{\text{ + }}\left( {{\text{ - 1}}} \right){\text{ + 5 }} $

$ {\text{ = - 1 + 1 - 1 + 5 = 4 }} $

Solve the following

(i) Add: ${\text{p (p - q), q (q - r)}}$ and ${\text{r (r - p)}}$

$ {\text{First expression = p }}\left( {{\text{p - q}}} \right){\text{ = }}{{\text{p}}^2}{\text{ - pq }} $

$ {\text{Second expression = q }}\left( {{\text{q - r}}} \right){\text{ = }}{{\text{q}}^2}{\text{ - qr}} $

$ {\text{Third expression = r }}\left( {{\text{r - p}}} \right){\text{ = }}{{\text{r}}^2}{\text{ - pr}} $

Adding the three expressions, we obtain

$ {\text{ }}{{\text{p}}^{\text{2}}}{\text{ - pq }} $

$ {\text{ + }}{{\text{q}}^{\text{2}}}{\text{ - qr}} $

$ {\text{ + }}{{\text{r}}^{\text{2}}}{\text{ - pr}} $

$ \overline {{\text{ }}{{\text{p}}^{\text{2}}}{\text{ - pq + }}{{\text{q}}^{\text{2}}}{\text{ - qr + }}{{\text{r}}^{\text{2}}}{\text{ - pr}}} $

Therefore, the sum is ${{\text{p}}^{\text{2}}}{\text{ - pq + }}{{\text{q}}^{\text{2}}}{\text{ - qr + }}{{\text{r}}^{\text{2}}}{\text{ - pr}}$

(ii) Add: ${\text{2x }}\left( {{\text{z - x - y}}} \right){\text{ and 2y }}\left( {{\text{z - y - x}}} \right){\text{ }}$

$ {\text{First expression = 2x }}\left( {{\text{z - x - y}}} \right){\text{ = 2xz - 2}}{{\text{x}}^{\text{2}}}{\text{ - 2xy }} $

$ {\text{Second expression = 2y }}\left( {{\text{z - y - x}}} \right){\text{ = 2yz - 2}}{{\text{y}}^{\text{2}}}{\text{ - 2yx }} $

Adding the two expressions, we obtain

$ {\text{ 2xz - 2}}{{\text{x}}^{\text{2}}}{\text{ - 2xy }} $

$ {\text{ + - 2yx + 2yz - 2}}{{\text{y}}^{\text{2}}}{\text{ }} $

$ \overline {\,\,{\text{ 2xz - 2}}{{\text{x}}^{\text{2}}}{\text{ - 4xy + 2yz - 2}}{{\text{y}}^{\text{2}}}} $

Therefore, the sum is ${\text{2xz - 2}}{{\text{x}}^{\text{2}}}{\text{ - 4xy + 2yz - 2}}{{\text{y}}^{\text{2}}}$

(iii) Subtract ${\text{3l }}\left( {{\text{l - 4m + 5n}}} \right){\text{ from 4l }}\left( {{\text{10n - 3m + 2l}}} \right){\text{ }}$

$ {\text{3l }}\left( {{\text{l - 4m + 5n}}} \right){\text{ = 3}}{{\text{l}}^{\text{2}}}{\text{ - 12lm + 15ln }} $

$ {\text{4l }}\left( {{\text{10n - 3m + 2l}}} \right){\text{ = 40ln - 12lm + 8}}{{\text{l}}^{\text{2}}}{\text{ }} $

Subtracting these expressions, we obtain

$ {\text{ 8}}{{\text{l}}^{\text{2}}}{\text{ - 12lm + 40ln}} $

$ {\text{ 3}}{{\text{l}}^{\text{2}}}{\text{ - 12lm + 15ln}} $

$ ( - )\,{\text{ }}( + ){\text{ }}( - ) $

$ \overline {{\text{ 5}}{{\text{l}}^2}{\text{ + 25ln }}} {\text{ }} $

Therefore, the result is ${\text{5}}{{\text{l}}^2}{\text{ + 25ln}}$

(iv) Subtract ${\text{3a }}\left( {{\text{a + b + c}}} \right){\text{ - 2b }}\left( {{\text{a - b + c}}} \right){\text{ from 4c }}\left( {{\text{ - a + b + c}}} \right)$

$ {\text{ + 4}}{{\text{c}}^{\text{2}}}{\text{ - 4ac + 4bc }} $

$ {\text{ 3}}{{\text{a}}^{\text{2}}}{\text{ + 2}}{{\text{b}}^{{\text{2 }}}}{\text{ + ab + 3ac - 2bc}} $

$ {\text{( - ) ( - ) ( - ) ( - ) ( + )}} $

$ \overline {{\text{ - 3}}{{\text{a}}^{\text{2}}}{\text{ - 2}}{{\text{b}}^{\text{2}}}\,{\text{ + 4}}{{\text{c}}^{\text{2}}}{\text{ + ab - 7ac + 6bc}}} $

Therefore, the result is ${\text{ - 3}}{{\text{a}}^{\text{2}}}{\text{ - 2}}{{\text{b}}^{\text{2}}}\,{\text{ + 4}}{{\text{c}}^{\text{2}}}{\text{ + ab - 7ac + 6bc}}$

Exercise – 9.4

1. Multiply the binomials.

(i) ${\text{(2x + 5)}}$and ${\text{(4x - 3)}}$

Ans: ${\text{(2x + 5) }} \times {\text{ (4x - 3) = 2x }} \times {\text{(4x - 3) + 5}} \times {\text{(4x - 3)}}$

${\text{ = 8}}{{\text{x}}^2}{\text{ - 6x + 20x - 15}}$

${\text{ = 8x2 + 14x - 15 (By adding like terms)}}$

(ii) ${\text{(y - 8)}}$and ${\text{(3y - 4)}}$

Ans: ${{ (y - 8) \times (3y - 4) = y \times (3y - 8) - 8 \times (3y - 4)}}$

${\text{ = 3}}{{\text{y}}^2}{\text{ - 4y - 24y + 32}}$

${\text{ = 3}}{{\text{y}}^{\text{2}}}{\text{ - 28y + 32 (By adding like terms)}}$

(iii) ${\text{(2}}{\text{.5l - 0}}{\text{.5m)}}$and ${\text{(2}}{\text{.5l + 0}}{\text{.5m)}}$

Ans: ${\text{(2}}{\text{.5l - 0}}{\text{.5m)(2}}{\text{.5l + 0}}{\text{.5m) = 2}}{{.5l \times (2}}{\text{.5l + 0}}{\text{.5m) - 0}}{\text{.5m(2}}{\text{.5l + 0}}{\text{.5m)}}$

${\text{ = 6}}{\text{.25}}{{\text{l}}^2}{\text{ + 1}}{\text{.25lm - 1}}{\text{.25lm - 0}}{\text{.25}}{{\text{m}}^2}$

${\text{ = 6}}{\text{.25}}{{\text{l}}^2}{\text{ - 0}}{\text{.25}}{{\text{m}}^2}$

(iv) $\left( {{\text{a + 3b}}} \right)$and ${\text{(x + 5)}}$

Ans: ${\text{(a + 3b) }} \times {\text{ (x + 5) = a}} \times {{(x + 5) + 3b }} \times {\text{(x + 5)}}$

${\text{ = ax + 5a + 3bx + 15b}}$

(v) ${\text{(2pq + 3}}{{\text{q}}^2}{\text{)}}$and ${\text{(3pq - 2}}{{\text{q}}^2}{\text{)}}$

Ans: ${\text{(2pq + 3}}{{\text{q}}^2}{\text{)}} \times {\text{(3pq - 2}}{{\text{q}}^2}{\text{) = 2pq }} \times {\text{(3pq - 2}}{{\text{q}}^2}{\text{) + 3}}{{\text{q}}^2} \times {\text{(3pq - 2}}{{\text{q}}^2}{\text{)}}$

${\text{ = 6p2}}{{\text{q}}^2}{\text{ - 4p}}{{\text{q}}^3}{\text{ + 9p}}{{\text{q}}^3}{\text{ - 6}}{{\text{q}}^4}$

${\text{ = 6p2}}{{\text{q}}^2}{\text{ + 5p}}{{\text{q}}^3}{\text{ - 6}}{{\text{q}}^4}$

(vi) $\left( {\dfrac{3}{4}{a^2} + 3{b^2}} \right)$and $4\left( {{a^2} - \dfrac{2}{3}{b^2}} \right)$

Ans: $\left( {\dfrac{{\text{3}}}{{\text{4}}}{{\text{a}}^{\text{2}}}{\text{ + 3}}{{\text{b}}^{\text{2}}}} \right){{ \times }}\left[ {{\text{4}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}\dfrac{{\text{2}}}{{\text{3}}}{{\text{b}}^{\text{2}}}} \right)} \right]{\text{ = }}\left( {\dfrac{{\text{3}}}{{\text{4}}}{{\text{a}}^{\text{2}}}{\text{ + 3}}{{\text{b}}^{\text{2}}}} \right){{ \times }}\left( {{\text{4}}{{\text{a}}^{\text{2}}}{\text{ - }}\dfrac{{\text{8}}}{{\text{3}}}{{\text{b}}^{\text{2}}}} \right)$

${\text{ = }}\dfrac{{\text{3}}}{{\text{4}}}{{\text{a}}^{\text{2}}}{{ \times }}\left( {{\text{4}}{{\text{\alpha }}^{\text{2}}}{\text{ - }}\dfrac{{\text{8}}}{{\text{3}}}{{\text{b}}^{\text{2}}}} \right){\text{ + 3}}{{\text{b}}^{\text{2}}}{{ \times }}\left( {{\text{4}}{{\text{a}}^{\text{2}}}{\text{ - }}\dfrac{{\text{8}}}{{\text{3}}}{{\text{b}}^{\text{2}}}} \right) $

$ {\text{ = 3}}{{\text{a}}^{\text{4}}}{\text{ - 2}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{\text{ + 12}}{{\text{b}}^{\text{2}}}{{\text{a}}^{\text{2}}}{\text{ - 8}}{{\text{b}}^{\text{4}}} $

$ {\text{ = 3}}{{\text{a}}^{\text{4}}}{\text{ + 10}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{\text{ - 8}}{{\text{b}}^{\text{4}}} $

2. Find the product.

(i) ${\text{(5 - 2x) (3 + x)}}$

Ans: ${\text{(5 - 2x) (3 + x) = 5 (3 + x) - 2x (3 + x)}}$

$ {\text{ = 15 + 5x - 6x - 2}}{{\text{x}}^2} $

$ {\text{ = 15 - x - 2}}{{\text{x}}^2} $

(ii) ${\text{(x + 7y) (7x - y)}}$

Ans: ${\text{(x + 7y) (7x - y) = x (7x - y) + 7y (7x - y)}}$

$ {\text{ = 7}}{{\text{x}}^2}{\text{ - xy + 49xy - 7}}{{\text{y}}^2} $

$ {\text{ = 7}}{{\text{x}}^2}{\text{ + 48xy - 7}}{{\text{y}}^2} $

(iii) ${\text{(}}{{\text{a}}^2}{\text{ + b) (a + }}{{\text{b}}^2}{\text{)}}$

Ans: ${\text{(}}{{\text{a}}^2}{\text{ + b) (a + }}{{\text{b}}^2}{\text{) = }}{{\text{a}}^2}{\text{ (a + }}{{\text{b}}^2}{\text{) + b (a + }}{{\text{b}}^2}{\text{)}}$

${\text{ = }}{{\text{a}}^3}{\text{ + }}{{\text{a}}^2}{{\text{b}}^2}{\text{ + ab + }}{{\text{b}}^3}$

(iv) ${\text{(}}{{\text{p}}^2}{\text{ - }}{{\text{q}}^2}{\text{) (2p + q)}}$

Ans: ${\text{(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0}}$

${\text{ = 2}}{{\text{p}}^3}{\text{ + }}{{\text{p}}^2}{\text{q - 2p}}{{\text{q}}^2}{\text{ - }}{{\text{q}}^3}$

3. Simplify.

(i) ${\text{(}}{{\text{x}}^2}{\text{ - 5) (x + 5) + 25}}$

Ans: ${\text{(}}{{\text{x}}^2}{\text{ - 5) (x + 5) + 25}}$

$ {{\text{x}}^2}{\text{ (x + 5) - 5 (x + 5) + 25}} $

$ {\text{ = }}{{\text{x}}^3}{\text{ + 5}}{{\text{x}}^2}{\text{ - 5x - 25 + 25}} $

$ {\text{ = }}{{\text{x}}^3}{\text{ + 5}}{{\text{x}}^2}{\text{ - 5x}} $

(ii) ${\text{(}}{{\text{a}}^2}{\text{ + 5) (}}{{\text{b}}^3}{\text{ + 3) + 5}}$

Ans: ${\text{(}}{{\text{a}}^2}{\text{ + 5) (}}{{\text{b}}^3}{\text{ + 3) + 5}}$

$ {\text{ = }}{{\text{a}}^2}{\text{ (}}{{\text{b}}^3}{\text{ + 3) + 5 (}}{{\text{b}}^3}{\text{ + 3) + 5}} $

$ {\text{ = }}{{\text{a}}^2}{{\text{b}}^3}{\text{ + 3}}{{\text{a}}^2}{\text{ + 5}}{{\text{b}}^3}{\text{ + 15 + 5}} $

$ {\text{ = }}{{\text{a}}^2}{{\text{b}}^3}{\text{ + 3}}{{\text{a}}^2}{\text{ + 5}}{{\text{b}}^3}{\text{ + 20}} $

(iii) ${\text{(t + }}{{\text{s}}^2}{\text{) (}}{{\text{t}}^2}{\text{ - s)}}$

Ans: ${\text{(t + }}{{\text{s}}^2}{\text{) (}}{{\text{t}}^2}{\text{ - s)}}$

$ {\text{ = t (}}{{\text{t}}^2}{\text{ - s) + }}{{\text{s}}^2}{\text{ (}}{{\text{t}}^2}{\text{ - s)}} $

$ {\text{ = }}{{\text{t}}^3}{\text{ - st + }}{{\text{s}}^2}{{\text{t}}^2}{\text{ - }}{{\text{s}}^3} $

(iv) ${\text{(a + b) (c - d) + (a - b) (c + d) + 2 (ac + bd)}}$

Ans: ${\text{(a + b) (c - d) + (a - b) (c + d) + 2 (ac + bd)}}$

$ {\text{ = a (c - d) + b (c - d) + a (c + d) - b (c + d) + 2 (ac + bd)}} $

$ {\text{ = ac - ad + bc - bd + ac + ad - bc - bd + 2ac + 2bd}} $

$ {\text{ = (ac + ac + 2ac) + (ad - ad) + (bc - bc) + (2bd - bd - bd)}} $

$ {\text{ = 4ac}} $

(v) ${\text{(x + y) (2x + y) + (x + 2y) (x - y)}}$

Ans: ${\text{(x + y) (2x + y) + (x + 2y) (x - y)}}$

$ {\text{ = x (2x + y) + y (2x + y) + x (x - y) + 2y (x - y)}} $

$ {\text{ = 2}}{{\text{x}}^2}{\text{ + xy + 2xy + }}{{\text{y}}^2}{\text{ + }}{{\text{x}}^2}{\text{ - xy + 2xy - 2}}{{\text{y}}^2} $

$ {\text{ = (2}}{{\text{x}}^2}{\text{ + }}{{\text{x}}^2}{\text{) + (}}{{\text{y}}^2}{\text{ - 2}}{{\text{y}}^2}{\text{) + (xy + 2xy - xy + 2xy)}} $

$ {\text{ = 3}}{{\text{x}}^2}{\text{ - }}{{\text{y}}^2}{\text{ + 4xy}} $

(vi) ${\text{(x + y) (}}{{\text{x}}^2}{\text{ - xy + }}{{\text{y}}^2}{\text{)}}$

Ans: ${\text{(x + y) (}}{{\text{x}}^2}{\text{ - xy + }}{{\text{y}}^2}{\text{)}}$

$ {\text{ = x (}}{{\text{x}}^2}{\text{ - xy + }}{{\text{y}}^2}{\text{) + y (}}{{\text{x}}^2}{\text{ - xy + }}{{\text{y}}^2}{\text{)}} $

$ {\text{ = }}{{\text{x}}^3}{\text{ - }}{{\text{x}}^2}{\text{y + x}}{{\text{y}}^2}{\text{ + }}{{\text{x}}^2}{\text{y - x}}{{\text{y}}^2}{\text{ + }}{{\text{y}}^3} $

$ {\text{ = }}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3}{\text{ + (x}}{{\text{y}}^2}{\text{ - x}}{{\text{y}}^2}{\text{) + (}}{{\text{x}}^2}{\text{y - }}{{\text{x}}^2}{\text{y)}} $

$ {\text{ = }}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3} $

(vii) ${\text{(1}}{\text{.5x - 4y) (1}}{\text{.5x + 4y + 3) - 4}}{\text{.5x + 12y}}$

Ans: ${\text{(1}}{\text{.5x - 4y) (1}}{\text{.5x + 4y + 3) - 4}}{\text{.5x + 12y}}$

$ {\text{ = 1}}{\text{.5x (1}}{\text{.5x + 4y + 3) - 4y (1}}{\text{.5x + 4y + 3) - 4}}{\text{.5x + 12y}} $

$ {\text{ = 2}}{\text{.25 }}{{\text{x}}^2}{\text{ + 6xy + 4}}{\text{.5x - 6xy - 16}}{{\text{y}}^2}{\text{ - 12y - 4}}{\text{.5x + 12y}} $

$ {\text{ = 2}}{\text{.25 }}{{\text{x}}^2}{\text{ + (6xy - 6xy) + (4}}{\text{.5x - 4}}{\text{.5x) - 16}}{{\text{y}}^2}{\text{ + (12y - 12y)}} $

$ {\text{ = 2}}{\text{.25}}{{\text{x}}^2}{\text{ - 16}}{{\text{y}}^2} $

(viii) ${\text{(a + b + c) (a + b - c)}}$

Ans: ${\text{(a + b + c) (a + b - c)}}$

$ {\text{ = a (a + b - c) + b (a + b - c) + c (a + b - c)}} $

$ {\text{ = }}{{\text{a}}^2}{\text{ + ab - ac + ab + }}{{\text{b}}^2}{\text{ - bc + ca + bc - }}{{\text{c}}^2} $

$ {\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ - }}{{\text{c}}^2}{\text{ + (ab + ab) + (bc - bc) + (ca - ca)}} $

$ {\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ - }}{{\text{c}}^2}{\text{ + 2ab}} $

Exercise – 9.5

Use a suitable identity to get each of the following products.

(i) ${\text{(x + 3) (x + 3) }}$

Ans: The products will be as follows.

$ {\text{ = (x}}{{\text{)}}^2}{\text{ + 2(x) (3) + (3}}{{\text{)}}^2}{\text{ [(a + b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + 2ab + }}{{\text{b}}^2}{\text{]}} $

$ {\text{ = }}{{\text{x}}^2}{\text{ + 6x + 9}} $

(ii) ${\text{(2y + 5) (2y + 5) }}$

${\text{(2y + 5) (2y + 5) = (2y + 5}}{{\text{)}}^2}$

$ {\text{ = (2y}}{{\text{)}}^2}{\text{ + 2(2y) (5) + (5}}{{\text{)}}^2}{\text{ [(a + b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + 2ab + }}{{\text{b}}^2}{\text{]}} $

$ {\text{ = 4}}{{\text{y}}^2}{\text{ + 20y + 25}} $

(iii) ${\text{(2a - 7) (2a - 7) }}$

${\text{(2a - 7) (2a - 7) = (2a - 7}}{{\text{)}}^2}$

${\text{[(a - b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ - 2ab + }}{{\text{b}}^2}{\text{]}}$

(iv) $\left( {3a - \dfrac{1}{2}} \right)\left( {3a - \dfrac{1}{2}} \right)$

Ans: $\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right) = {\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)^2}$

${\left( {{\text{3a}}} \right)^{\text{2}}}{\text{ - 2(3a)}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ + }}{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)^{\text{2}}}$

${\text{ = 9}}{{\text{a}}^{\text{2}}}{\text{ - 3a + }}\dfrac{{\text{1}}}{{\text{4}}}$

(v) ${\text{(1}}{\text{.1m - 0}}{\text{.4) (1}}{\text{.1 m + 0}}{\text{.4)}}$

Ans: ${\text{(1}}{\text{.1m - 0}}{\text{.4) (1}}{\text{.1 m + 0}}{\text{.4)}}$

$ {\text{ = (1}}{\text{.1m}}{{\text{)}}^2}{\text{ - (0}}{\text{.4}}{{\text{)}}^2}{\text{ [(a + b) (a - b) = }}{{\text{a}}^2}{\text{ - }}{{\text{b}}^2}{\text{]}} $

$ {\text{ = 1}}{\text{.21}}{{\text{m}}^2}{\text{ - 0}}{\text{.16}} $

(vi) ${\text{(}}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{) ( - }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{)}}$

Ans: ${\text{(}}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{) ( - }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{)}}$

$ {\text{ = (}}{{\text{b}}^2}{{\text{)}}^2}{\text{ - (}}{{\text{a}}^2}{{\text{)}}^2}{\text{ [(a + b) (a - b) = }}{{\text{a}}^2}{\text{ - }}{{\text{b}}^2}{\text{]}} $

$ {\text{ = }}{{\text{b}}^4}{\text{ - }}{{\text{a}}^4} $

(vii) ${\text{(6x - 7) (6x + 7)}}$

Ans: ${\text{(6x - 7) (6x + 7) = (6x}}{{\text{)}}^2}{\text{ - (7}}{{\text{)}}^2}{\text{ [(a + b) (a - b) = }}{{\text{a}}^2}{\text{ - }}{{\text{b}}^2}{\text{]}}$

${\text{ = 36}}{{\text{x}}^2}{\text{ - 49}}$

(viii) ${\text{( - a + c) ( - a + c)}}$

Ans: ${\text{( - a + c) ( - a + c) = ( - a + c}}{{\text{)}}^2}$

$ {\text{ = ( - a}}{{\text{)}}^2}{\text{ + 2( - a) (c) + (c}}{{\text{)}}^2}{\text{ [(a + b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + 2ab + }}{{\text{b}}^2}{\text{]}} $

$ {\text{ = }}{{\text{a}}^2}{\text{ - 2ac + }}{{\text{c}}^2} $

(ix) $\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)$

Ans: $\left( {\dfrac{{\text{x}}}{{\text{2}}}{\text{ + }}\dfrac{{{\text{3y}}}}{{\text{4}}}} \right)\left( {\dfrac{{\text{x}}}{{\text{2}}}{\text{ + }}\dfrac{{{\text{3y}}}}{{\text{4}}}} \right){\text{ = }}{\left( {\dfrac{{\text{x}}}{{\text{2}}}{\text{ + }}\dfrac{{{\text{3y}}}}{{\text{4}}}} \right)^{\text{2}}}$

${\text{ = 16}}{{\text{x}}^2}{\text{ + 24x + 5}}$ ${\text{[(a + b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + 2ab + }}{{\text{b}}^2}{\text{]}}$

${\dfrac{{\text{x}}}{{\text{4}}}^{\text{2}}}{\text{ + }}\dfrac{{{\text{3xy}}}}{{\text{4}}}{\text{ + }}\dfrac{{{\text{9}}{{\text{y}}^{\text{2}}}}}{{{\text{16}}}}$

(x) ${\text{(7a - 9b) (7a - 9b)}}$

Ans: ${\text{(7a - 9b) (7a - 9b) = (7a - 9b}}{{\text{)}}^2}$

$ {\text{ = (7a}}{{\text{)}}^2}{\text{ - 2(7a)(9b) + (9b}}{{\text{)}}^2}{\text{ [(a - b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ - 2ab + }}{{\text{b}}^2}{\text{]}} $

$ {\text{ = 49}}{{\text{a}}^2}{\text{ - 126ab + 81}}{{\text{b}}^2} $

2. Use the identity ${\text{(x + a) (x + b) = }}{{\text{x}}^2}{\text{ + (a + b)x + ab}}$ to find the following products.

(i) ${\text{(x + 3) (x + 7)}}$

Ans: ${\text{(x + 3) (x + 7) = }}{{\text{x}}^2}{\text{ + (3 + 7) x + (3) (7)}}$

${\text{ = }}{{\text{x}}^2}{\text{ + 10x + 21}}$

(ii) ${\text{(4x + 5) (4x + 1) }}$

Ans: ${\text{(4x + 5) (4x + 1) = (4x}}{{\text{)}}^2}{\text{ + (5 + 1) (4x) + (5) (1)}}$

${\text{ = 16}}{{\text{x}}^2}{\text{ + 24x + 5}}$

(iii) ${\text{(4x + 5) (4x - 1) }}$

Ans: ${\text{(4x - 5) (4x - 1) = (4x}}{{\text{)}}^2}{\text{ + }}\left[ {{\text{( - 5) + ( - 1)}}} \right]{\text{ (4x) + ( - 5) ( - 1)}}$

$ {\text{ = (}}{{\text{m}}^2}{{\text{)}}^2}{\text{ - 2(}}{{\text{m}}^2}{\text{) (}}{{\text{n}}^2}{\text{m) + (}}{{\text{n}}^2}{\text{m}}{{\text{)}}^2}{\text{ + 2}}{{\text{m}}^3}{{\text{n}}^2}{\text{ [(a - b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ - 2ab + }}{{\text{b}}^2}{\text{ ]}} $

$ {\text{ = }}{{\text{m}}^4}{\text{ - 2}}{{\text{m}}^3}{{\text{n}}^2}{\text{ + }}{{\text{n}}^4}{{\text{m}}^2}{\text{ + 2}}{{\text{m}}^3}{{\text{n}}^2} $

$ {\text{ = }}{{\text{m}}^4}{\text{ + }}{{\text{n}}^4}{{\text{m}}^2} $

(iv) ${\text{(4x + 5) (4x - 1) }}$

Ans: ${\text{(4x + 5) (4x - 1) = (4x}}{{\text{)}}^2}{\text{ + }}\left[ {{\text{(5) + ( - 1)}}} \right]{\text{ (4x) + (5) ( - 1)}}$

${\text{ = 16}}{{\text{x}}^2}{\text{ + 16x - 5}}$

(v) ${\text{(2x + 5y) (2x + 3y)}}$

Ans: ${\text{(2x + 5y) (2x + 3y) = (2x}}{{\text{)}}^2}{\text{ + (5y + 3y) (2x) + (5y) (3y)}}$

${\text{ = 4}}{{\text{x}}^2}{\text{ + 16xy + 15}}{{\text{y}}^2}$

(vi) ${\text{(2}}{{\text{a}}^2}{\text{ + 9) (2}}{{\text{a}}^2}{\text{ + 5)}}$

Ans: ${\text{(2}}{{\text{a}}^2}{\text{ + 9) (2}}{{\text{a}}^2}{\text{ + 5) = (2}}{{\text{a}}^2}{{\text{)}}^2}{\text{ + (9 + 5) (2}}{{\text{a}}^2}{\text{) + (9) (5)}}$

${\text{ = 4}}{{\text{a}}^4}{\text{ + 28}}{{\text{a}}^2}{\text{ + 45}}$

(vii) ${\text{(xyz - 4) (xyz - 2)}}$

Ans: ${{\text{(xyz)}}^2}{\text{ + }}\left[ {( - 4) + ( - 2)} \right]{\text{ (xyz) + ( - 4) ( - 2)}}$

${\text{ = }}{{\text{x}}^2}{{\text{y}}^2}{{\text{z}}^2}{\text{ - 6xyz + 8}}$

Find the following squares by suing the identities

(i) ${{\text{(b - 7)}}^2}$

Ans: ${{\text{(b - 7)}}^2}{\text{ = (b}}{{\text{)}}^2}{\text{ - 2(b) (7) + (7}}{{\text{)}}^2}{\text{ [(a - b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ - 2ab + }}{{\text{b}}^2}{\text{]}}$

${\text{ = }}{{\text{b}}^2}{\text{ - 14b + 49}}$

(ii) ${{\text{(xy + 3z)}}^2}$

Ans: ${{\text{(xy + 3z)}}^2}{\text{ = (xy}}{{\text{)}}^2}{\text{ + 2(xy) (3z) + (3z}}{{\text{)}}^2}{\text{ [(a + b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + 2ab + }}{{\text{b}}^2}{\text{]}}$

${\text{ = }}{{\text{x}}^2}{{\text{y}}^2}{\text{ + 6xyz + 9}}{{\text{z}}^2}$

(iii) ${{\text{(6}}{{\text{x}}^2}{\text{ - 5y)}}^2}$

Ans: ${\text{ = (6}}{{\text{x}}^2}{{\text{)}}^2}{\text{ - 2(6}}{{\text{x}}^2}{\text{) (5y) + (5y}}{{\text{)}}^2}{\text{ [(a - b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ - 2ab + }}{{\text{b}}^2}{\text{]}}$

${\text{ = 36}}{{\text{x}}^4}{\text{ - 60}}{{\text{x}}^2}{\text{y + 25}}{{\text{y}}^2}$

(iv) ${\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{m + }}\dfrac{{\text{3}}}{{\text{2}}}{\text{n}}} \right)^{\text{2}}}$

Ans: ${\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{m + }}\dfrac{{\text{3}}}{{\text{2}}}{\text{n}}} \right)^{\text{2}}}{\text{ = }}{\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{m}}} \right)^{\text{2}}}{\text{ + 2}}\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{m}}} \right)\left( {\dfrac{{\text{3}}}{{\text{2}}}{\text{n}}} \right){\text{ + }}{\left( {\dfrac{{\text{3}}}{{\text{2}}}{\text{n}}} \right)^{\text{2}}}\left[ {{{{\text{(a + b)}}}^{\text{2}}}{\text{ = }}{{\text{a}}^{\text{2}}}{\text{ + 2ab + }}{{\text{b}}^{\text{2}}}} \right]$

${\text{ = }}\dfrac{{\text{4}}}{{\text{9}}}{{\text{m}}^{\text{2}}}{\text{ + 2mn + }}\dfrac{{\text{9}}}{{\text{4}}}{{\text{n}}^{\text{2}}}$

(v) ${{\text{(0}}{\text{.4p - 0}}{\text{.5q)}}^2}$

Ans: ${{\text{(0}}{\text{.4p - 0}}{\text{.5q)}}^2}{\text{ = (0}}{\text{.4p}}{{\text{)}}^2}{\text{ - 2 (0}}{\text{.4p) (0}}{\text{.5q) + (0}}{\text{.5q}}{{\text{)}}^2}$

$ {\text{[(a - b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ - 2ab + }}{{\text{b}}^2}{\text{]}} $

$ {\text{ = 0}}{\text{.16}}{{\text{p}}^2}{\text{ - 0}}{\text{.4pq + 0}}{\text{.25}}{{\text{q}}^2} $

(vi) ${{\text{(2xy + 5y)}}^2}$

Ans: ${{\text{(2xy + 5y)}}^2}{\text{ = (2xy}}{{\text{)}}^2}{\text{ + 2(2xy) (5y) + (5y}}{{\text{)}}^2}$

$ {\text{[(a + b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + 2ab + }}{{\text{b}}^2}{\text{]}} $

$ {\text{ = 4}}{{\text{x}}^2}{{\text{y}}^2}{\text{ + 20x}}{{\text{y}}^2}{\text{ + 25}}{{\text{y}}^2} $

(i) ${{\text{(}}{{\text{a}}^2}{\text{ - }}{{\text{b}}^2}{\text{)}}^2}$

Ans: ${{\text{(}}{{\text{a}}^2}{\text{ - }}{{\text{b}}^2}{\text{)}}^2}{\text{ = (}}{{\text{a}}^2}{{\text{)}}^2}{\text{ - 2(}}{{\text{a}}^2}{\text{) (}}{{\text{b}}^2}{\text{) + (}}{{\text{b}}^2}{{\text{)}}^2}{\text{ }}$${\text{[(a - b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ - 2ab + }}{{\text{b}}^2}{\text{ ]}}$

${\text{ = }}{{\text{a}}^4}{\text{ - 2}}{{\text{a}}^2}{{\text{b}}^2}{\text{ + }}{{\text{b}}^4}$

(ii) ${{\text{(2x + 5)}}^2}{\text{ - (2x - 5}}{{\text{)}}^2}{\text{ }}$

Ans: ${{\text{(2x + 5)}}^2}{\text{ - (2x - 5}}{{\text{)}}^2}{\text{ = (2x}}{{\text{)}}^2}{\text{ + 2(2x) (5) + (5}}{{\text{)}}^2}{\text{ - [(2x}}{{\text{)}}^2}{\text{ - 2(2x) (5) + (5}}{{\text{)}}^2}{\text{]}}$

$ {\text{ = 4}}{{\text{x}}^2}{\text{ + 20x + 25 - [4}}{{\text{x}}^2}{\text{ - 20x + 25]}} $

$ {\text{ = 4}}{{\text{x}}^2}{\text{ + 20x + 25 - 4}}{{\text{x}}^2}{\text{ + 20x - 25 = 40x}} $

(iii) ${{\text{(7m - 8n)}}^2}{\text{ + (7m + 8n}}{{\text{)}}^2}$

Ans: ${{\text{(7m - 8n)}}^2}{\text{ + (7m + 8n}}{{\text{)}}^2}$

$ {\text{ = (7m}}{{\text{)}}^2}{\text{ - 2(7m) (8n) + (8n}}{{\text{)}}^2}{\text{ + (7m}}{{\text{)}}^2}{\text{ + 2(7m) (8n) + (8n}}{{\text{)}}^2} $

$ {\text{[(a - b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ - 2ab + }}{{\text{b}}^2}{\text{ and (a + b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + 2ab + }}{{\text{b}}^2}{\text{]}} $

$ {\text{ = 49}}{{\text{m}}^2}{\text{ - 112mn + 64}}{{\text{n}}^2}{\text{ + 49}}{{\text{m}}^2}{\text{ + 112mn + 64}}{{\text{n}}^2} $

$ {\text{ = 98}}{{\text{m}}^2}{\text{ + 128}}{{\text{n}}^2} $

(iv) ${{\text{(4m + 5n)}}^2}{\text{ + (5m + 4n}}{{\text{)}}^2}$

Ans: ${{\text{(4m + 5n)}}^2}{\text{ + (5m + 4n}}{{\text{)}}^2}$

$ {\text{ = (4m}}{{\text{)}}^2}{\text{ + 2(4m) (5n) + (5n}}{{\text{)}}^2}{\text{ + (5m}}{{\text{)}}^2}{\text{ + 2(5m) (4n) + (4n}}{{\text{)}}^2} $

$ {\text{[ (a + b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + 2ab + }}{{\text{b}}^2}{\text{]}} $

$ {\text{ = 16}}{{\text{m}}^2}{\text{ + 40mn + 25}}{{\text{n}}^1}{\text{ + 25}}{{\text{m}}^2}{\text{ + 40mn + 16}}{{\text{n}}^2} $

$ {\text{ = 41}}{{\text{m}}^2}{\text{ + 80mn + 41}}{{\text{n}}^2} $

(v) ${{\text{(2}}{\text{.5p - 1}}{\text{.5q)}}^2}{\text{ - (1}}{\text{.5p - 2}}{\text{.5q}}{{\text{)}}^2}$

Ans: ${{\text{(2}}{\text{.5p - 1}}{\text{.5q)}}^2}{\text{ - (1}}{\text{.5p - 2}}{\text{.5q}}{{\text{)}}^2}$

$ {\text{ = (2}}{\text{.5p}}{{\text{)}}^2}{\text{ - 2(2}}{\text{.5p) (1}}{\text{.5q) + (1}}{\text{.5q}}{{\text{)}}^2}{\text{ - [(1}}{\text{.5p}}{{\text{)}}^2}{\text{ - 2(1}}{\text{.5p)(2}}{\text{.5q) + (2}}{\text{.5q}}{{\text{)}}^2}{\text{]}} $

$ {\text{[(a - b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ - 2ab + }}{{\text{b}}^2}{\text{ ]}} $

$ {\text{ = 6}}{\text{.25}}{{\text{p}}^2}{\text{ - 7}}{\text{.5pq + 2}}{\text{.25}}{{\text{q}}^2}{\text{ - [2}}{\text{.25}}{{\text{p}}^2}{\text{ - 7}}{\text{.5pq + 6}}{\text{.25}}{{\text{q}}^2}{\text{]}} $

$ {\text{ = 6}}{\text{.25}}{{\text{p}}^2}{\text{ - 7}}{\text{.5pq + 2}}{\text{.25}}{{\text{q}}^2}{\text{ - 2}}{\text{.25}}{{\text{p}}^2}{\text{ + 7}}{\text{.5pq - 6}}{\text{.25}}{{\text{q}}^2}{\text{]}} $

$ {\text{ = 4}}{{\text{p}}^2}{\text{ - 4}}{{\text{q}}^2} $

(vi) ${{\text{(ab + bc)}}^2}{\text{ - 2a}}{{\text{b}}^2}{\text{c}}$

Ans: ${{\text{(ab + bc)}}^2}{\text{ - 2a}}{{\text{b}}^2}{\text{c}}$

$ {\text{ = (ab}}{{\text{)}}^2}{\text{ + 2(ab)(bc) + (bc}}{{\text{)}}^2}{\text{ - 2a}}{{\text{b}}^2}{\text{c [(a + b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + 2ab + }}{{\text{b}}^2}{\text{ ]}} $

$ {\text{ = }}{{\text{a}}^2}{{\text{b}}^2}{\text{ + 2a}}{{\text{b}}^2}{\text{c + }}{{\text{b}}^2}{{\text{c}}^2}{\text{ - 2a}}{{\text{b}}^2}{\text{c}} $

${\text{ = }}{{\text{a}}^2}{{\text{b}}^2}{\text{ + }}{{\text{b}}^2}{{\text{c}}^2} $

(vii) ${{\text{(}}{{\text{m}}^2}{\text{ - }}{{\text{n}}^2}{\text{m)}}^2}{\text{ + 2}}{{\text{m}}^3}{{\text{n}}^2}$

Ans: ${{\text{(}}{{\text{m}}^2}{\text{ - }}{{\text{n}}^2}{\text{m)}}^2}{\text{ + 2}}{{\text{m}}^3}{{\text{n}}^2}$

$ {\text{ = }}{{\text{m}}^4}{\text{ + }}{{\text{n}}^4}{{\text{m}}^2} $

(i) ${{\text{(3x + 7)}}^2}{\text{ - 84x = (3x - 7}}{{\text{)}}^2}$

Ans: Taking L.H.S.

${{\text{(3x + 7)}}^2}{\text{ - 84x}}$

$ {\text{ = (3x}}{{\text{)}}^2}{\text{ + 2(3x)(7) + (7}}{{\text{)}}^2}{\text{ - 84x}} $

$ {\text{ = 9}}{{\text{x}}^2}{\text{ + 42x + 49 - 84x}} $

$ {\text{ = 9}}{{\text{x}}^2}{\text{ - 42x + 49}} $

Now considering R.H.S.

$ {\text{ = (3x - 7}}{{\text{)}}^2}{\text{ = (3x}}{{\text{)}}^2}{\text{ - 2(3x)(7) + (7}}{{\text{)}}^2} $

Therefore, L.H.S. = R.H.S.

(ii) ${{\text{(9p - 5q)}}^2}{\text{ + 180pq = (9p + 5q}}{{\text{)}}^2}$

${{\text{(9p - 5q)}}^2}{\text{ + 180pq}}$

$ {\text{ = (9p}}{{\text{)}}^2}{\text{ - 2(9p)(5q) + (5q}}{{\text{)}}^2}{\text{ - 180pq}} $

$ {\text{ = 81}}{{\text{p}}^2}{\text{ - 90pq + 25}}{{\text{q}}^2}{\text{ + 180pq}} $

$ {\text{ = 81}}{{\text{p}}^2}{\text{ + 90pq + 25}}{{\text{q}}^2} $

Now, considering R.H.S.

${{\text{(9p + 5q)}}^2}$

$ {\text{ = (9p}}{{\text{)}}^2}{\text{ + 2(9p)(5q) + (5q}}{{\text{)}}^2} $

(iii) ${\left( {\dfrac{{\text{4}}}{{\text{3}}}{\text{m - }}\dfrac{{\text{3}}}{{\text{4}}}{\text{n}}} \right)^{\text{2}}}{\text{ + 2mn = }}\dfrac{{{\text{16}}}}{{\text{9}}}{{\text{m}}^{\text{2}}}{\text{ + }}\dfrac{{\text{9}}}{{{\text{16}}}}{{\text{n}}^{\text{2}}}$

${\left( {\dfrac{{\text{4}}}{{\text{3}}}{\text{m - }}\dfrac{{\text{3}}}{{\text{4}}}{\text{n}}} \right)^{\text{2}}}$

$ {\text{ = }}{\left( {\dfrac{{\text{4}}}{{\text{3}}}{\text{m}}} \right)^{\text{2}}}{\text{ - 2}}\left( {\dfrac{{\text{4}}}{{\text{3}}}{\text{m}}} \right)\left( {\dfrac{{\text{3}}}{{\text{4}}}{\text{n}}} \right){\text{ + }}{\left( {\dfrac{{\text{3}}}{{\text{4}}}{\text{n}}} \right)^{\text{2}}}{\text{ + 2mn}} $

$ {\text{ = }}\dfrac{{{\text{16}}}}{{\text{9}}}{\text{m}}{{\text{r}}^{\text{2}}}{\text{ - 2mn + }}\dfrac{{\text{9}}}{{{\text{16}}}}{{\text{n}}^{\text{2}}}{\text{ + 2mn}} $

$ {\text{ = }}\dfrac{{{\text{16}}}}{{\text{9}}}{{\text{m}}^{\text{2}}}{\text{ + }}\dfrac{{\text{9}}}{{{\text{16}}}}{{\text{n}}^{\text{2}}}{\text{ = R}}{\text{.H}}{\text{.S}} $

Therefore, L.H.S.=R.H.S.

(iv) ${{\text{(4pq + 3q)}}^2}{\text{ - (4pq - 3q}}{{\text{)}}^2}{\text{ = 48p}}{{\text{q}}^2}$

$ {\text{ = (4pq}}{{\text{)}}^2}{\text{ + 2(4pq)(3q) + (3q}}{{\text{)}}^2}{\text{ - [(4pq}}{{\text{)}}^2}{\text{ - 2(4pq) (3q) + (3q}}{{\text{)}}^2}{\text{]}} $

$ {\text{ = 16}}{{\text{p}}^2}{{\text{q}}^2}{\text{ + 24p}}{{\text{q}}^2}{\text{ + 9}}{{\text{q}}^2}{\text{ - [16}}{{\text{p}}^2}{{\text{q}}^2}{\text{ - 24p}}{{\text{q}}^2}{\text{ + 9}}{{\text{q}}^2}{\text{]}} $

$ {\text{ = 16}}{{\text{p}}^2}{{\text{q}}^2}{\text{ + 24p}}{{\text{q}}^2}{\text{ + 9}}{{\text{q}}^2}{\text{ - 16}}{{\text{p}}^2}{{\text{q}}^2}{\text{ + 24p}}{{\text{q}}^2}{\text{ - 9}}{{\text{q}}^2} $

$ {\text{ = 48p}}{{\text{q}}^2}{\text{ = R}}{\text{.H}}{\text{.S}} $

(v) ${\text{(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0}}$

Ans: Taking L.H.S

${\text{(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a)}}$

${\text{ = (}}{{\text{a}}^2}{\text{ - }}{{\text{b}}^2}{\text{) + (}}{{\text{b}}^2}{\text{ - }}{{\text{c}}^2}{\text{) + (}}{{\text{c}}^2}{\text{ - }}{{\text{a}}^2}{\text{) = 0 = R}}{\text{.H}}{\text{.S}}{\text{. = 0}}$

6. Using identities, evaluate .

(i) ${\text{7}}{{\text{1}}^2}$

Ans: ${\text{7}}{{\text{1}}^2}{\text{ = (70 + 1}}{{\text{)}}^2}$ ${\text{[(a + b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + 2ab + }}{{\text{b}}^2}{\text{ ]}}$

$ {\text{ = (70}}{{\text{)}}^2}{\text{ + 2(70) (1) + (1}}{{\text{)}}^2}{\text{ }} $

$ {\text{ = 4900 + 140 + 1 = 5041}} $

(ii) ${\text{9}}{{\text{9}}^2}$

Ans: ${\text{9}}{{\text{9}}^2}{\text{ = (100 - 1}}{{\text{)}}^2}$ ${\text{[(a - b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ - 2ab + }}{{\text{b}}^2}{\text{ ]}}$

${\text{ = (100}}{{\text{)}}^2}{\text{ - 2(100) (1) + (1}}{{\text{)}}^2}{\text{ }}$

$ {\text{ = 10000 - 200 + 1 = 9801}} $

(iii) ${\text{10}}{{\text{2}}^2}{\text{ }}$

Ans: ${\text{10}}{{\text{2}}^2}{\text{ = (100 + 2}}{{\text{)}}^2}$ ${\text{[(a + b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + 2ab + }}{{\text{b}}^2}{\text{ ]}}$

$ {\text{ = (100}}{{\text{)}}^2}{\text{ + 2(100)(2) + (2}}{{\text{)}}^2} $

$ {\text{ = 10000 + 400 + 4 = 10404}} $

(iv) ${\text{99}}{{\text{8}}^2}$

Ans: ${\text{99}}{{\text{8}}^2}{\text{ = (1000 - 2}}{{\text{)}}^2}$ ${\text{[(a + b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + 2ab + }}{{\text{b}}^2}{\text{ ]}}$

$ {\text{ = (1000}}{{\text{)}}^2}{\text{ - 2(1000)(2) + (2}}{{\text{)}}^2} $

$ {\text{ = 1000000 - 4000 + 4 = 996004}} $

(v) ${{\text{(5}}{\text{.2)}}^2}{\text{ }}$

Ans: ${{\text{(5}}{\text{.2)}}^2}{\text{ = (5}}{\text{.0 + 0}}{\text{.2}}{{\text{)}}^2}$ ${\text{[(a + b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + 2ab + }}{{\text{b}}^2}{\text{ ]}}$

$ {\text{ = (5}}{\text{.0}}{{\text{)}}^2}{\text{ + 2(5}}{\text{.0) (0}}{\text{.2) + (0}}{\text{.2}}{{\text{)}}^2}{\text{ }} $

$ {\text{ = 25 + 2 + 0}}{\text{.04 = 27}}{\text{.04}} $

(vi) ${\text{297 }} \times {\text{ 303}}$

Ans: ${\text{297 }} \times {\text{ 303 = (300 - 3) }} \times {\text{ (300 + 3) }}$ [ ${{\text{a}}^2}{\text{ - }}{{\text{b}}^2}{\text{ = (a + b) (a - b),}}$ ]

${\text{ = (300}}{{\text{)}}^2}{\text{ - (3}}{{\text{)}}^2}{\text{ }}$

${\text{ = 90000 - 9 = 89991}}$

(vii) ${\text{78 }} \times {\text{ 82}}$

Ans: ${\text{78 }} \times {\text{ 82 = (80 - 2) (80 + 2)}}$ [ ${{\text{a}}^2}{\text{ - }}{{\text{b}}^2}{\text{ = (a + b) (a - b),}}$ ]

$ {\text{ = (80}}{{\text{)}}^2}{\text{ - (2}}{{\text{)}}^2}{\text{ }} $

$ {\text{ = 6400 - 4 = 6396}} $

(viii) ${\text{8}}{\text{.}}{{\text{9}}^2}{\text{ }}$

Ans: ${\text{8}}{\text{.}}{{\text{9}}^2}{\text{ = (9}}{\text{.0 - 0}}{\text{.1}}{{\text{)}}^2}$ ${\text{[(a - b}}{{\text{)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ - 2ab + }}{{\text{b}}^2}{\text{ ]}}$

$ {\text{ = (9}}{\text{.0}}{{\text{)}}^2}{\text{ - 2(9}}{\text{.0) (0}}{\text{.1) + (0}}{\text{.1}}{{\text{)}}^2} $

$ {\text{ = 81 - 1}}{\text{.8 + 0}}{\text{.01 = 79}}{\text{.21}} $

(ix) ${\text{1}}{\text{.05 }} \times \,{\text{9}}{\text{.5}}$

Ans: ${\text{1}}{\text{.05 }} \times \,{\text{9}}{\text{.5 = 1}}{\text{.05 }} \times 0.95 \times 10$

${\text{ = (1 + 0}}{\text{.05) (1 - 0}}{\text{.05) }}$ [ ${{\text{a}}^2}{\text{ - }}{{\text{b}}^2}{\text{ = (a + b) (a - b),}}$

${\text{ = [(1}}{{\text{)}}^2}{\text{ - (0}}{\text{.05}}{{\text{)}}^2}{\text{] }}$

${\text{ = [1 - 0}}{\text{.0025] }} \times {\text{ 10}}$

${\text{ = 0}}{\text{.9975 }} \times \,{\text{10 = 9}}{\text{.975}}$

Using ${{\text{a}}^2}{\text{ - }}{{\text{b}}^2}{\text{ = (a + b) (a - b),}}$ find

(i) ${\text{5}}{{\text{1}}^2}{\text{ - 4}}{{\text{9}}^2}$

Ans: ${\text{5}}{{\text{1}}^2}{\text{ - 4}}{{\text{9}}^2}$

${\text{5}}{{\text{1}}^2}{\text{ - 4}}{{\text{9}}^2}{\text{ = (51 + 49) (51 - 49)}}$

${\text{ = (100) (2) = 200}}$

(ii) ${{\text{(1}}{\text{.02)}}^2}{\text{ - (0}}{\text{.98}}{{\text{)}}^2}{\text{ }}$

Ans: ${{\text{(1}}{\text{.02)}}^2}{\text{ - (0}}{\text{.98}}{{\text{)}}^2}{\text{ }}$

${{\text{(1}}{\text{.02)}}^2}{\text{ - (0}}{\text{.98}}{{\text{)}}^2}{\text{ = (1}}{\text{.02 + 0}}{\text{.98) (1}}{\text{.02 - 0}}{\text{.98)}}$

(iii) ${\text{15}}{{\text{3}}^2}{\text{ - 14}}{{\text{7}}^2}$

Ans: ${\text{15}}{{\text{3}}^2}{\text{ - 14}}{{\text{7}}^2}$

${\text{15}}{{\text{3}}^2}{\text{ - 14}}{{\text{7}}^2}{\text{ = (153 + 147) (153 - 147)}}$

${\text{ = (300) (6) = 1800}}$

(iv) ${\text{12}}{\text{.}}{{\text{1}}^2}{\text{ - 7}}{\text{.}}{{\text{9}}^2}$

Ans: ${\text{12}}{\text{.}}{{\text{1}}^2}{\text{ - 7}}{\text{.}}{{\text{9}}^2}$

${\text{12}}{\text{.}}{{\text{1}}^2}{\text{ - 7}}{\text{.}}{{\text{9}}^2}{\text{ = (12}}{\text{.1 + 7}}{\text{.9) (12}}{\text{.1 - 7}}{\text{.9)}}$

8. Using ${\text{(x + a) (x + b) = }}{{\text{x}}^2}{\text{ + (a + b) x + ab,}}$ find

(i) ${\text{103 }} \times \,{\text{104}}$

Ans: ${\text{103 }} \times \,{\text{104 = (100 + 3) (100 + 4)}}$

$ {\text{ = (100}}{{\text{)}}^2}{\text{ + (3 + 4) (100) + (3) (4)}} $

$ {\text{ = 10000 + 700 + 12 = 10712}} $

(ii) ${\text{5}}{\text{.1 }} \times {\text{ 5}}{\text{.2}}$

Ans: ${\text{5}}{\text{.1 }} \times {\text{ 5}}{\text{.2 = (5 + 0}}{\text{.1) (5 + 0}}{\text{.2)}}$

$ {\text{ = (5}}{{\text{)}}^2}{\text{ + (0}}{\text{.1 + 0}}{\text{.2) (5) + (0}}{\text{.1) (0}}{\text{.2)}} $

$ {\text{ = 25 + 1}}{\text{.5 + 0}}{\text{.02 = 26}}{\text{.52}} $

(iii) ${\text{103 }} \times \,{\text{98}}$

Ans: ${\text{103 }} \times \,{\text{98 = (100 + 3) (100 - 2)}}$

$ {\text{ = (100}}{{\text{)}}^2}{\text{ + [3 + ( - 2)] (100) + (3) ( - 2)}} $

$ {\text{ = 10000 + 100 - 6}} $

$ {\text{ = 10094}} $

(iv) ${\text{9}}{\text{.7 }} \times \,\,9.8$

Ans: ${\text{9}}{\text{.7 }} \times \,\,9.8{\text{ = (10 - 0}}{\text{.3) (10 - 0}}{\text{.2)}}$

$ {\text{ = (10}}{{\text{)}}^2}{\text{ + [( - 0}}{\text{.3) + ( - 0}}{\text{.2)] (10) + ( - 0}}{\text{.3) ( - 0}}{\text{.2)}} $

$ {\text{ = 100 + ( - 0}}{\text{.5)10 + 0}}{\text{.06 = 100}}{\text{.06 - 5 = 95}}{\text{.06}}

## NCERT Solutions for Class 8 Maths Chapter 9 - Free PDF Download

If you are worried about internet connectivity then don’t worry as NCERT Solutions Class 8 Maths Chapter 9 are available in pdf format. They are easy to download and after downloading, these solutions can be accessed as per students' wish. These NCERT Solutions Class 8 are available on our website and our app. NCERT Solutions Class 8 is entirely free of cost. So if you are going to have a test or exam near, our NCERT Solutions Class 8 is there for you. These solutions help in the last minute revision of the chapters and ensure that one does not miss the most important questions of NCERT.

## NCERT Solutions for Class 8 Maths Chapter 9

Chapter - 9 algebraic expressions and identities.

In the curriculum of Class 8 Maths Chapter 9 is Algebraic Expressions and Identities.

Algebra is introduced in Class 8th to make the students know about the algebraic terms, algebraic identities, variables, constants, algebraic expressions, monomial, binomial and trinomial expressions. Also, the mathematical operations like addition, subtraction, multiplication, and division of algebraic expressions are explained in this chapter.

Algebra and its real-life applications are an important part of the maths syllabus. Various Exercises are given in Chapter 9 Maths for students to solve and become familiar with algebra.

With our Grade 8 NCERT solutions, it becomes easy for students to understand the algebraic concepts and solve the questions. These solutions are prepared by our subject experts.

## Class 8 Maths Chapter 9 Marks Weightage

Chapter 9 Algebraic Expressions and Identities is a very important chapter from any exam point of view whether it's school exams or competitive exams thus going through these NCERT Solutions will help the student to get a good score on their exams.

In this chapter, a total of 5 Exercises are given with different types of questions and our solutions will help the students to solve these questions.

## We Cover all Exercises in the Chapter Given Below:

Why are ncert solutions class 8 chapter 9 important.

Preparing from our NCERT Solutions Class 8 helps students to gain confidence over the chapter through detailed explanations given in our NCERT solutions.

NCERT solutions ensure that students have gone through all the questions given in NCERT which are important from an exam point of view.

Last-minute revision and preparation can be easily done by these solutions.

NCERT Solutions are made in simple language which makes it understandable for the students.

NCERT solutions also give the students an idea of how a mathematical solution must be written.

## NCERT Class 8 Maths Chapter wise Solutions in Hindi

Chapter 1 - Rational Numbers in Hindi

Chapter 2 - Linear Equations in One Variable in Hindi

Chapter 3 - Understanding Quadrilaterals in Hindi

Chapter 4 - Practical Geometry in Hindi

Chapter 5 - Data Handling in Hindi

Chapter 6 - Squares and Square Roots in Hindi

Chapter 7 - Cubes and Cube Roots in Hindi

Chapter 8 - Comparing Quantities in Hindi

Chapter 9 - Algebraic Expressions and Identities in Hindi

Chapter 10 - Visualising Solid Shapes in Hindi

Chapter 11 - Mensuration in Hindi

Chapter 12 - Exponents and Powers in Hindi

Chapter 13 - Direct and Inverse Proportions in Hindi

Chapter 14 - Factorisation in Hindi

Chapter 15 - Introduction to Graphs in Hindi

Chapter 16 - Playing with Number in Hindi

## NCERT Solutions for Class 8 Maths - Chapterwise Solutions

Chapter 1 - Rational Numbers

Chapter 2 - Linear Equations in One Variable

Chapter 3 - Understanding Quadrilaterals

Chapter 4 - Practical Geometry

Chapter 5 - Data Handling

Chapter 6 - Squares and Square Roots

Chapter 7 - Cubes and Cube Roots

Chapter 8 - Comparing Quantities

Chapter 9 - Algebraic Expressions and Identities

Chapter 10 - Visualising Solid Shapes

Chapter 11 - Mensuration

Chapter 12 - Exponents and Powers

Chapter 13 - Direct and Inverse Proportions

Chapter 14 - Factorisation

Chapter 15 - Introduction to Graphs

Chapter 16 - Playing with Numbers

## Conclusion

The NCERT Solutions for Class 8 Maths Chapter 9 on Algebraic Expressions and Identities by Vedantu are a helpful resource for students. This chapter focuses on understanding and working with algebraic expressions, which are mathematical phrases involving numbers, variables, and operations. It's an important foundation for higher-level math. One crucial section is likely the exploration of identities, where students learn about equations that hold true for any value of the variables. Mastering this concept is key for solving complex mathematical problems. Vedantu's solutions provide clear explanations, making the learning process more accessible and supporting students in grasping the fundamentals of algebra.

## FAQs on NCERT Solutions for Class 8 Maths Chapter 9 - Algebraic Expressions And Identities

Q1.What are the Topics Covered in the Class 8 Chapter 9 Maths?

Ans: Chapter 9 of Grade 8 Maths is Algebraic Expressions and Identities.

The topics covered in this chapter in the NCERT book are as follows:

Meaning of algebraic expressions

What are the terms, factors, and coefficients

What are monomials, binomials, and polynomials

Like and Unlike terms

Addition and subtraction of algebraic terms

Multiplication of algebraic terms

Identities, standard identities, and application of identities.

All these topics are the core topics covered. Subtopics like multiplication of monomial to monomial, polynomial to polynomial, monomial to polynomial are also covered in the chapter.

Q2. Are these NCERT Solutions Helpful in Scoring Good Marks in School Exams?

Ans: Our 8th standard Maths algebra NCERT solutions are prepared by experts who are highly qualified and experienced. These solutions are prepared in such a manner that the methods and mathematical steps can be easily understood by the students. When students can gain the concept they will be able to solve the questions related to algebra that may come in their school exams and this will increase their score in their exams. All the Exercises given in the NCERT book in Chapter 9 are included in our algebraic expressions NCERT Class 8 solutions.

Q3. What are algebraic expressions in Class 8?

Ans: In Class 8, algebraic expressions are equations or expressions formed by the combination of variables, numbers and algebraic operations (addition, subtraction, multiplication etc.). An Algebraic expression is a term consisting of variables, coefficients and a constant. For example: 4x + 7y - 2. Here, 4 and 7 are coefficients, x and y are variables, and ‘2’ is a constant. Algebraic expressions are of three types - monomial, binomial, and trinomial.

Q4. What is the formula for algebraic expression?

Ans: Class 8 Maths introduces the concept of algebraic expressions and algebraic identities to the students. Algebraic identities are algebraic equations that are valid for all variable values in them. Following are the formulas for algebraic expressions:

(a+b) 2 = a 2 +2ab +b 2

(a-b) 2 = a 2 -2ab +b 2

(a+b)(a-b)= a 2 -b 2

(x+a)(x+b)= x 2 +(a+b) x+ab

(x+a)(x-b)= x 2 +(a-b) x-ab

(x-a)(x+b)= x 2 +(b-a) x-ab

(x-a)(x-b)= x 2 -(a+b) x+ab

(a+b) 3 = a 3 +3ab(a+b) +b 3

(a-b) 3 = a 3 -3ab(a-b) -b 3

You can learn more about algebraic formulas and their applications from Vedantu.

Q5. How can I prepare myself to score good marks in Maths Class 8?

Ans: To score good grades in Maths Class 8, students should refer to the study material available on Vedantu. Vedantu provides you with notes and NCERT Solutions. In these solutions, all the exercises from the Class 8 Maths NCERT textbook have been thoroughly addressed. Also, NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities are available on the Vedantu app for free of cost. Furthermore, these solutions might help you prepare for a variety of competitive exams.

Q6. Where can I get NCERT Solutions for Class 8 Maths Chapter 9?

Ans: Students can find the NCERT Solutions for Class 8 Maths Chapter 9 “Algebraic Expressions and Identities” on Vedantu. The NCERT Solutions provided by Vedantu are the best because of the accuracy and precision with which they are prepared. Vedantu offers free chapter-by-chapter NCERT Solutions for Class 8 Maths in PDF format. These solutions are created by subject matter experts with years of experience.

Q7. What concepts can I learn using the NCERT Solutions for Class 8 Maths Chapter 9?

Ans: NCERT Solutions for Class 8 Maths Chapter 9 provided in the website and the app of Vedantu ensure to clear all the concepts given in the NCERT Maths Textbook Class 8. It provides an in-depth explanation of what an Algebraic expression is and what algebraic identities are. It provides students with not only the theoretical part but also the sample questions for practice.

## NCERT Solutions for Class 8 Maths

Ncert solutions for class 8.

- Class 8 Maths MCQs
- Chapter 9 Algebraic Expression and Identities

## Class 8 Maths Chapter 9 Algebraic Expressions and Identities MCQs

Class 8 Maths Chapter 9 Algebraic Expressions and Identities MCQs (Questions and Answers) are provided online here for students. These multiple-choice questions have been designed by experts referring to the latest CBSE syllabus (2022-2023) and NCERT guidelines. The objective questions are provided chapter-wise at BYJU’S. Also, learn important questions for class 8 Maths here.

Practice more and test your skills on Class 8 Maths Chapter 9 Algebraic Expressions and Identities MCQs with the given PDF here.

## MCQs Questions on Class 8 Algebraic Expressions and Identities

Multiple choice questions (MCQs) are available for Class 8 chapter 9 Algebraic Expressions and Identities chapter. For all the problems, there are four multiple options given among which one is the right answer. Solve each question and choose the correct answer.

1. An algebraic expression that contains only one term is called:

A. Monomial

B. Binomial

C. Trinomial

D. None of the above

Example: 2x is a monomial

2. 5x+6y is a:

Explanation: The expression containing two terms is called binomial.

3. The algebraic expression 3x+2y+6 is a:

Explanation: The algebraic expression containing three terms is called a trinomial. Here, 3x, 2y and 6 are three terms.

4. A polynomial contains _______ number of terms:

Explanation: A polynomial can contain any number of terms, i.e. one or more than one.

5. In which of the following, the two expressions are like terms?

A. 7x and 7y

B. 7x and 9x

C. 7x and 7x 2

D. 7x and 7xy

6. If we add, 7xy + 5yz – 3zx, 4yz + 9zx – 4y and –3xz + 5x – 2xy, then the answer is:

A. 5xy + 9yz +3zx + 5x – 4y

B. 5xy – 9yz +3zx – 5x – 4y

C. 5xy + 10yz +3zx + 15x – 4y

D. 5xy + 10yz +3zx + 5x – 6y

Explanation: Given, 7xy + 5yz – 3zx, 4yz + 9zx – 4y and –3xz + 5x – 2xy.

If we add the three expressions, then we need to combine the like terms together.

(7xy – 2xy) + (5yz + 4yz) – 3zx + 9zx – 3xz – 4y + 5x

= 5xy + 9yz + 3zx + 5x – 4y

7. If we subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3, then the answer is:

A. 8a+2ab+2b+15

B. 8a+2ab+2b-15

C. 8a-2ab+2b-15

D. 8a-2ab-2b-15

Explanation: (12a – 9ab + 5b – 3) – (4a – 7ab + 3b + 12)

= 12a – 9ab + 5b – 3 – 4a + 7ab -3b-12

= (12 – 4)a – (9 – 7)ab + (5 – 3)b – 3 – 12

= 8a – 2ab + 2b – 15

8. If we multiply 5x and (– 4xyz), then we get:

A. 20x 2 yz

B. -20x 2 yz

Explanation: (5x) x (-4xyz)

= 5 × x × (-4) × x × y × z

= -20x 1+1 yz

= -20x 2 yz

9. The product of 4x and 0 is:

Explanation: Any value multiplied by zero is zero.

10. The volume of a cuboid with length, breadth and height as 5x, 3x 2 and 7x 4 respectively is:

Explanation: Volume of cuboid = Length × breadth × height

V = 5x × 3x 2 × 7x 4

V = 105 x 1+2+4

V = 105x 7 cubic units

11. The product of 5x and 3y is:

Answer: (D) 15xy

(5x)(3y) = 15xy

12. The product of 6x and -11x is:

Answer: B. -66x²

(6x) (-11x) = –66x²

13. The area of a rectangle whose length and breadth are 3y and 9y² respectively is:

Answer: C. 27y³

Explanation:

Area of rectangle = length x breadth = 3y x 9y² = 27y³.

14. The area of a rectangle that has length = 2a²b and breadth = 3ab² is:

Answer: A. 6a³b³

Area of rectangle = (2a²b)(3ab²) = 6a³b³

15. The side of a cube is 2a. Find the volume of the cube.

Answer: C. 8a³

Volume of the cube = 2a × 2a × 2a = 8a³

16. Multiplication of monomials x², (–x)³, (–x) 4 is equal to:

Answer: A. x 9

Explanation: (x²).(-x³).(-x) 4 = x 9

17. The value of (x – y)(x + y) + (y – z)(y + z) + (z – x) (z + x) is:

A. x + y + z

B. x² + y² + z²

C. xy + yz + zx

Answer: D. 0

(x – y)(x + y) + (y – z)(y + z) + (z – x) (z + x)

18. (a – b)² is equal to:

A. a² + b² – 2ab

B. a² + b² + 2ab

Answer: A. a² + b² – 2ab

Explanation: By algebraic identity,

(a + b)² = a² + b² – 2 ab

19. The product of 3xy 2 z and 4x is:

C. 12x 2 y 2 z

D. 12x 2 yz

Answer: C. 12x 2 y 2 z

Explanation: The product of 3xy 2 z and 4x is:

⇒ (3xy 2 z) (4x)

⇒ 3.x.y 2 .z.4.x

⇒ 12x 2 y 2 z

20. Which of the following is a like term as 8xy?

Answer: D. xy

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## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. To score good marks in Class 8 Mathematics examination, it is advised to solve questions provided at the end of each chapter in the NCERT book.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities are prepared based on Class 8 NCERT syllabus, taking the types of questions asked in the NCERT textbook into consideration. Further, all the CBSE Class 8 Solutions Maths Chapter 9 are in accordance with the latest CBSE guidelines and marking schemes.

## Class 8 Maths Chapter 9 Exercise 9.1 Solutions

## Class 8 Maths Chapter 9 Exercise 9.2 Solutions

## Class 8 Maths Chapter 9 Exercise 9.3 Solutions

## Class 8 Maths Chapter 9 Exercise 9.4 Solutions

## Class 8 Maths Chapter 9 Exercise 9.5 Solutions

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## NCERT Solutions for Class 8 Maths Algebraic Expressions and Identities

NCERT Solutions for Class 8 Maths Algebraic Expressions and Identities are created by askIITians experts to help you understand all the basic concepts of the chapter. With these solutions, you can easily revise the main topics of the chapter and prepare yourself for the Class 8 final Maths examination. We have provided stepwise solutions for all the exercise questions of the chapter with comprehensive explanations. You can download the NCERT Solutions for this chapter for free in pdf form and practice its exercises whenever you want.

Chapter 9 Algebraic Expressions and Identities is an important chapter as it helps build a fundamental understanding of important concepts of algebra. The chapter includes basic definitions of expressions, terms, variables and coefficients. You will study different types of polynomial expressions such as monomials, binomials, trinomials, along with like terms, unlike terms, and addition, subtraction and multiplication of algebraic expressions.

## About NCERT Class 8 Maths Chapter 9 Algebraic Expressions and Identities

- Algebraic Expression: Any mathematical expression which consists of numbers, variables and operations is called an Algebraic Expression . For example, 7x + 4, 22y + 91
- Terms: Every algebraic expression is made up of two or more terms . Terms are added to form expressions. Terms themselves can be formed as the product of factors. For example, the term 4x is the product of its factors 4 and x. The term 5 is made up of just one factor, i.e., 5.
- Monomial: An algebraic expression with only one term is called a monomial. For example, 2x, 76y, x 2 y, etc are monomials.
- Polynomial: Algebraic expressions that have more than two terms with non-zero coefficients and variables having non-negative integral exponents are called polynomials. For example, x+y+z, 2x+45y, etc are polynomials.
- Binomial: An algebraic expression that includes only two terms is known as a binomial. For example, 5a 2 + 5, 3x+3y etc are binomial expressions.

(a + b) 2 = a 2 + 2ab + b 2

(a - b) 2 = a 2 - 2ab + b 2

(a + b) (a - b) = a 2 - b 2

## Download Free NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

At askIITians, you can find exercise-wise solutions for Chapter 9 Algebraic Expressions and Identities. This way you can practise the exercises at your own pace and learn independently. Our comprehensive solutions will make it easier for you to understand the important concepts of the chapter and prepare well for your exams.

Chapter 9 Class 8 Maths NCERT Ex 9.1 Solutions

The first exercise of Algebraic Expressions and Identities includes 4 questions. This exercise is based on the introduction to algebraic expressions, terms, coefficients, monomials, binomials, and polynomial expressions. It also includes questions related to like and unlike terms and addition and subtraction of algebraic expressions.

Chapter 9 Class 8 Maths NCERT Ex 9.2 Solutions

The second exercise of Algebraic Expressions and Identities includes 5 questions. This exercise is based on the introduction to the multiplication of algebraic expressions, multiplying a monomial by a monomial and multiplying three or more monomials. Make sure you practice this exercise carefully.

Chapter 9 Class 8 Maths NCERT Ex 9.3 Solutions

The third exercise of this chapter includes 5 questions. This exercise is based on multiplying a monomial by a binomial and multiplying a monomial by a trinomial. This exercise is based on exercise 9.2. So if you have understood the previous exercise, you will be able to solve its questions easily. In case you have any doubt, check our NCERT Solutions.

Chapter 9 Class 8 Maths NCERT Ex 9.4 Solutions

The fourth exercise of this chapter includes 3 questions. It is based on two important multiplication concepts - multiplication of a binomial by a binomial and multiplication of a binomial by a trinomial. This is an advanced level exercise. You must solve this exercise carefully as many students make mistakes in multiplying larger algebraic expressions.

Chapter 9 Class 8 Maths NCERT Ex 9.5 Solutions

The fifth exercise of this chapter includes 8 questions. It is based on important algebraic identities and their application. This exercise forms the basis of algebra in higher classes. So, you must practice every question of this exercise thoroughly. Our NCERT Solutions are here to guide you in case you have any doubts.

Why download NCERT Class 8 Maths Solutions for Chapter 9 Algebraic Expressions and Identities:

- NCERT Solutions are free to download in pdf format so that you can refer to them whenever you need them.
- The solutions include comprehensive explanations that will help in enhancing your conceptual understanding of the chapter.
- Every question is solved step by step which will make exam preparation easier for you.
- You can study the NCERT Solutions independently and practice the questions at your own pace.
- With these NCERT Solutions, you will understand how to approach different types of questions in the exams related to algebraic expressions and identities.

## NCERT Class 8 Maths Chapter 9 Algebraic Expressions and Identities FAQs

- How many questions are there in NCERT Class 8 Maths Chapter 9 Algebraic Expressions and Identities?

There are 25 unsolved questions in NCERT Class 8 Maths Chapter 9 Algebraic Expressions and Identities. You can download the stepwise solutions for all these questions from the askIITians website.

- I need more help in learning Chapter 9 Algebraic Expressions and Identities. What should I do?

You can find many different study resources for Class 8 Maths NCERT Chapter 9 Algebraic Expressions and Identities such as revision notes, mind maps, flashcards, extra questions, and more. We also provide online coaching for Class 8 Maths so that students can study from the best teachers of India through the safety and comfort of their homes.

- What are the topics covered in NCERT Class 8 Maths Chapter 9 Algebraic Expressions and Identities?

Class 8 Maths Chapter 9 includes the basic definitions of algebraic expressions, terms, variables and coefficients. It also includes different types of polynomial expressions such as monomials, binomials, trinomials, along with like terms, unlike terms, and addition, subtraction and multiplication of algebraic expressions.

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## MCQ Questions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities with Answers

We have compiled the NCERT MCQ Questions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities with Answers Pdf free download covering the entire syllabus. Practice MCQ Questions for Class 8 Maths with Answers on a daily basis and score well in exams. Refer to the Algebraic Expressions and Identities Class 8 MCQs Questions with Answers here along with a detailed explanation.

## Algebraic Expressions and Identities Class 8 MCQs Questions with Answers

Choose the correct option.

Question 1. Which of the following is the numeral coefficient of – 3x²y²? (a) 4 (b) 3 (c) 2 (d) -3

Answer: (d) -3

Question 2. What kind of polynomial is -5abc? (a) Monomial (b) Binomial (c) Trinonial (d) None of the above

Answer: (a) Monomial

Question 3. The value ofx² – 2yx + y² when x= 1; y = 2 is (a) 1 (b) 2 (c) 4 (d) 5

Answer: (a) 1

Question 4. How many terms are there in the expression 7 – 3x²y? (a) 1 (b) 2 (c) 3 (d) 4

Answer: (b) 2

Question 5. Which of the following is a binomial? (a) 3xy (b) 41 + 5m (c) 2x + 3y-5 (d) 4a – 7ab + 3b + 12

Answer: (b) 41 + 5m

Question 6. The sum of -7x, 3x and 11x (a) – 7x (b) 21x (c) 7x (d) -21x

Answer: (c) 7x

Question 7. What do you get when you subtract. -3xy from 5xy? (a) 3xy (b) 5xy (c) 8xy (d) 2xy

Answer: (c) 8xy

Question 8. The area of a rectangle with length 2l²m and breadth 31m² is (a) l³m³ (b) 2l³m³ (c) 4l³m³ (d) 6l³m³

Answer: (d) 6l³m³

Question 9. The volume of the cuboid of dimensions x, y, z is (a) x²y²z² (b) x³y³z³ (c) xyz (d) x²y²z

Answer: (b) x³y³z³

Question 10. The value of (x + y) (x + y) + (y – z) (y + z) + (z – x) (z + x) is equal to (a) 3x² (b) 3y² (c) 3z² (d) 0

Answer: (d) 0

Question 11. Which of the following is the numerical coefficient of x² y²? (a) 0 (b) 1 (c) x² (d) y²

Answer: (b) 1

Question 12. Which of the following is the numerical coefficient of -5xy? (a) 5 (b) x (c) 5 (d) y

Answer: (c) 5

Fill in the blanks

Question 1. The coefficient of-5x is ………………

Question 2. An expression having only one term is called ………………

Answer: monomial

Question 3. An expression having only three terms is called ………………

Answer: trinomial

Question 4. The product of (4x + y) (4x – y) is ………………….

Answer: 16x² – y²

Question 5. The sum of a a – b + 3 and 3a + 2b + 5 is ……………….

Answer: 4a + b + 8

Hope the information shed above regarding NCERT MCQ Questions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 8 Maths Algebraic Expressions and Identities MCQs Multiple Choice Questions with Answers, feel free to reach us so that we can revert back to us at the earliest possible.

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## CBSE Class 11 Applied Maths Algebra 2024: Syllabus & Prep Tips

Author : Tanya Kaushal

Updated On : May 30, 2024

Reader's Digest: Ready to learn the ins and outs of CBSE Class 11 Applied Maths Coordinate Geometry 2024? Read this blog to learn the concept, project topics, syllabus, weightage & prep tips!

The world of mathematics continues to evolve, as does its importance in many professions and daily applications. Recognizing this, the board announced the introduction of a new topic, Applied Mathematics.

A significant chapter from the CBSE class 11 Applied Mathematics is ' Algebra ', a cornerstone in mathematics. Delving deeper into the weightage, the algebra unit comprises 15 marks out of the 80 in the theory exam.

Here are the points to be discussed in this blog:

- Basics of CBSE Class 11 Applied Maths Coordinate Geometry: Dive into the foundational concepts and unique approach that sets Applied Maths Coordinate Geometry apart.
- Important Topics of CBSE Applied Maths Coordinate Geometry Class 11: Identify the core topics that deserve your focus and attention to excel in this subject.
- Table of Common Errors to Avoid: Uncover students' pitfalls and learn how to avoid them, ensuring a smoother learning experience.
- CBSE Class 11 Applied Maths Coordinate Geometry Solved Sample Questions: Put your knowledge to the test with solved sample questions that showcase the challenges you'll encounter in this exciting field.

Download Free Study Material for CBSE and CUET Exam 2024 by SuperGrads

## What is CBSE Class 11 Applied Maths Algebra?

Algebra is a branch of mathematics in which letters and symbols represent numbers and quantities in formulas and equations.

It provides methods to work with finite and infinite dimensional vector spaces and linear transformations between these spaces, including systems of linear equations, matrices, determinants, polynomials, and certain abstract structures, such as groups, rings, and fields.

Essentially, algebra provides a powerful tool to analyze and understand various mathematical structures.

- Linear Equations: If we're told that the sum of two numbers is 15 and one number is 5 greater than the other, we can use algebra to represent this situation. Let the smaller number be x. The other number is x + 5. Thus, the equation becomes x+x+5=15x+x+5=15. Solving this, we get x=5, so the numbers are 5 and 10.
- Polynomials: Consider the expression 2x² - 3x - 5. This is a polynomial of degree 2. When we set this equal to zero, we get a quadratic equation, which can be solved using various methods such as factoring, using the quadratic formula, or completing the square.

We can use methods like substitution, elimination, or matrix techniques to find the values of x and y that satisfy both equations simultaneously.

## What Are the Most Important Topics of CBSE Class 11 Applied Maths Algebra?

Applied mathematics course prepares you to choose algebraic methods as a means of representation and as a problem-solving tool. Algebra mainly focuses on topics like sets, relations, Venn diagrams, the relation between arithmetic and geometric progression, etc.

Go through the table below to learn the detailed CBSE Class 11 Applied Maths Syllabus for algebra .

## List of CBSE Class 11 Applied Maths Algebra Formulae

Algebra is the study of unknown quantities. Some of the topics of algebraic expressions and formulae are:

## Algebraic Identities

Various equality equations consist of different variables in algebraic identities.

- Linear Equations in One Variable: A linear equation in one variable has the maximum of one variable in order 1. It is depicted as ax + b = 0, where x represents the variable.
- Linear Equations in Two Variables: A linear equation in two variables consists of the utmost two variables present in order 2. The equation is depicted as ax2 + bx + c = 0. The two variables are crucial because your coursebook has a lot of questions based on it. So, you must stay focused on important algebra formulas to find the solution.

Some basic identities to note are:

- The combination of literal numbers obeys every fundamental rule of addition, subtraction, multiplication and division.
- x × y = xy; such as 5 × a = 5a = a × 5.
- a × a × a × … 9 more times = a12
- If a number is x8, then x is the base, and 8 is the exponent.
- A constant is a symbol with a fixed numerical value.

## Law of Exponent

The degrees and powers in any mathematical expression are known as exponents. Some of the laws of exponent are:

- (am)n = amn
- am / an = am-n
- am x bm = (ab)m
- am / bm = (a/b)m
- (a/b)-m =(b/a)m
- (1) n = 1 for infinite values of n

Read More: CBSE Class 11 Applied Maths - Mathematical Reasoning

## Quadratic Equations

The linear equations in two variables are known as quadratic equations.

The roots of the equation ax2 + bx + c = 0 (where a ≠ 0) can be given as:

− b ± b 2−4 ac √2 a

Below are some important points about the equation as a part of essential algebra formulas:

- Δ = b2 − 4ac is also known as a discriminant.

Delta; > 0 happens when the roots are real and distinct

For real and coincident roots, Δ = 0

Delta; < 0 happens in the case when the roots are non-real

- If α and β are the two roots of the equation ax2 + bx + c, then, α + β = (-b / a) and α × β = (c / a).
- If the roots of a quadratic equation are α and β, the equation will be (x − α)(x − β) = 0.

The general algebra formulas can be given as:

- n is a natural number: an – bn = (a – b)(an-1 + an-2b+…+ bn-2a + bn-1)
- If n is even: (n = 2k), an + bn = (a – b)(an-1 + an-2b +…+ bn-2a + bn-1)
- n is odd: (n = 2k + 1), an + bn = (a + b)(an-1 – an-2b +an-3b2…- bn-2a + bn-1)
- General square Formula: (a + b + c + …)2 = a2 + b2 + c2 + … + 2(ab + ac + bc + ….)

List of important formulas

- (a + b)2 = a2 + 2ab + b2
- (a – b)2 = a2 – 2ab + b2
- (a + b) (a – b) = a2 -b2
- (x + a) (x + b) = x2 + (a + b) x + ab
- (x + a) (x – b) = x2 + (a – b) x – ab
- (x – a) (x + b) = x2 + (b – a) x – ab
- (x – a) (x – b) = x2 – (a + b) x + ab
- (a + b)3 = a3 + b3 + 3ab (a + b)
- (a – b)3 = a3 – b3 – 3ab (a – b)
- (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
- (a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4
- (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz + 2xz
- (x + y – z)2 = x2 + y2 + z2 + 2xy – 2yz – 2xz
- (x – y + z)2 = x2 + y2 + z2 – 2xy – 2yz + 2xz
- (x – y – z)2 = x2 + y2 + z2 – 2xy + 2yz – 2xz
- x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz -xz)
- x2 + y2 = 12
- [(x + y)2 + (x – y)2]
- (x + a) (x + b) (x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
- x3 + y3 = (x + y) (x2 – xy + y2)
- x3 – y3 = (x – y) (x2 + xy + y2)
- x2 + y2 + z2 – xy – yz – z [(x – y)2 + (y – z)2 + (z – x)2]

Read More: CBSE Class 11 Applied Math Books

## What are the Common Mistakes to Avoid While Solving CBSE Class 11 Applied Maths Algebra Questions?

Here's a table detailing common errors students make while solving CBSE Class 11 Applied Maths Algebra problems and tips to avoid those errors:

Read More: CBSE Class 11 Applied Maths Calculus

## CBSE Class 11 Applied Maths Algebra Sample Questions

To ease your preparation, we have provided important topic-wise questions for class 11 Applied Maths Algebra in the post below.

Question 1: Write the following sets in the roaster form.

(i) A = {x | x is a positive integer less than 10 and 2x – 1 is an odd number}

(ii) C = {x : x2 + 7x – 8 = 0, x ∈ R}

Question 2: Write the following sets in roster form:

(i) A = {x : x is an integer and –3 ≤ x < 7}

(ii) B = {x : x is a natural number less than 6}

## Types of Sets

Question 1: Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, 0 < x < 10}; B = {2, 3, 5, 7}. Write the set (A U B)’.

Question 2: Let U = {1, 2, 3, 4, 5, 6, 7}, A = {2, 4, 6}, B = {3, 5} and C = {1, 2, 4, 7}, find

(i) A′ ∪ (B ∩ C′)

(ii) (B – A) ∪ (A – C)

## Venn Diagram

Question 1:

In a class of 50 students, 10 take Guitar lessons, 20 take singing classes, and 4 take both. Find the number of students who don’t take either Guitar or singing lessons.

## De Morgan's Laws

Prove De Morgan’s Laws by Venn Diagram

(i) (A∪B)’= A’∩ B’

## Relations & Types of Relations

Write the range of a Signup function.

Question 2:

The Cartesian product A × A has 9 elements, found (–1, 0) and (0,1). Find the set A and the remaining aspects of A × A.

## Introduction of Sequences, Series

Question 1: The sum of n terms of two arithmetic progressions is 5n+4: 9n+6. Find the ratio of their 18th terms.

Question 2: Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

## Arithmetic and Geometric Progression

Question 1: Find the sum of divisible integers from 1 to 100 by 2 or 5.

Question 2: What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

Check Out - CBSE Class 11 Numerical Applications

## Relationship Between AM and GM

Question 1: Find the AM, GM, and HM between 12 and 30

## Basic Concepts of Permutations and Combinations

Find the 3-digit numbers that can be formed from the given digits: 1, 2, 3, 4 and 5, assuming that

- digits can be repeated.
- Digits are not allowed to be repeated.

A coin is tossed 6 times, and the outcomes are noted. How many possible results can be there?

## Permutations, Circular Permutations, Permutations with restrictions

From a team of 6 students, in how many ways can we choose a captain and vice-captain, assuming one person cannot hold more than one position?

Question 2: Find the number of ways in which 10 beads can be arranged to form a necklace.

Question 3: Find the number of ways in which four girls and three boys can arrange themselves in a row so that none of the boys is together. How is this arrangement different from that in a circular way?

## Combinations with Standard Results

Question 1: How many words can be formed each of 2 vowels and 3 consonants from the letters of the given word – DAUGHTER?

Question 2: Find the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

Check Out - CBSE Class 11 Commerce Books

## CBSE Class 11 Applied Maths Algebra Sample Questions with Solutions

To help you understand the type of questions that can be asked in the exam, we have provided some sample questions for your reference here.

Question 1: How many words can be formed each of 2 vowels and 3 consonants from the given word's letters – DAUGHTER?

No. of Vowels in the word – DAUGHTER is 3.

No. of Consonants in the word Daughter is 5.

No of ways to select a vowel = 3c2 = 3!/2!(3 – 2)! = 3

No. of ways to select a consonant = 5c3 = 5!/3!(5 – 3)! = 10

Now you know that the number of combinations of 3 consonants and 2 vowels = 10x 3 = 30

Total number of words = 30 x 5! = 3600 ways.

Question 2: It is needed to seat 5 boys and 4 girls in a row to get the even places. How many such arrangements are possible?

5 boys and 4 girls are to be seated in a row to get the even places.

The 5 boys can be seated in 5! Ways.

For each of the arrangement, the 4 girls can be seated only at the places which are cross marked to make girls occupy the even places).

B x B x B x B x B

So, the girls can be seated in 4! Ways.

Hence, the possible number of arrangements = 4! × 5! = 24 × 120 = 2880

Question 3: Find the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

Take a deck of 52 cards,

To get exactly one king, 5-card combinations have to be made. It should be made in such a way that in each selection of 5 cards, or a deck of 52 cards, there will be 4 kings.

To select 1 king out of 4 kings = 4c1

To select 4 cards out of the remaining 48 cards = 48c4

To get the needed number of 5 card combination = 4c1 x 48c4

= 4x2x 47x 46×45

= 778320 ways.

Check Also: CBSE Class 11 Applied Maths Probability

Question 4 : Find the number of 6 digit numbers that can be formed by using the digits 0, 1, 3, 5, 7, and 9. These digits shall be divisible by 10, and no digit shall be repeated?

The number which has a 0 in its unit place is divisible by 10.

If we put 0 in the unit place, _ _ _ _ 0, there will be as many ways to fill 5 vacant places. (1, 3, 5, 7, 9)

The five vacant places can be filled in 5! ways = 120.

Question 5 : Evaluate 10! – 6!

10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x1 = 3628800

6! = 6 X 5 x 4 x 3 x 2 x 1 = 720

10! – 6! = 3628800 – 720 = 3628080

Question 6: Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.

Let a and d be the first term and the common difference of the A.P. respectively. It is known

that the kth term of an A.P. is given by

ak = a +(k -1)d

Therefore, am+n = a +(m+n -1)d

am-n = a +(m-n -1)d

am = a +(m-1)d

Hence, the sum of (m + n)th and (m – n)th terms of an A.P is written as:

am+n+ am-n = a +(m+n -1)d + a +(m-n -1)d

= 2a +(m + n -1+ m – n -1)d

=2a+(2m-2)d

=2a + 2(m-1)d

= 2 [a + (m-1)d]

= 2 am [since am = a +(m-1)d]

Therefore, the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.

Question 7: Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

The integers from 1 to 100, which are divisible by 2, are 2, 4, 6 ….. 100.

This forms an A.P. with both the first term and common difference equal to 2.

⇒ 100=2+(n-1)2

⇒ n= 50

Therefore, the sum of integers from 1 to 100 that are divisible by 2 is given as:

2+4+6+…+100 = (50/2)[2(2)+(50-1)(2)]

= (50/2)(4+98)

The integers from 1 to 100, which are divisible by 5, 10…. 100

This forms an A.P. with both the first term and common difference equal to 5.

Therefore, 100= 5+(n-1)5

⇒5n = 100

⇒ n= 100/5

⇒ n= 20

5+10+15+…+100= (20/2)[2(5)+(20-1)(5)]

= (20/2)(10+95)

Hence, the integers from 1 to 100, which are divisible by both 2 and 5 are 10, 20, ….. 100.

This also forms an A.P. with both the first term and common difference equal to 10.

Therefore, 100= 10+(n-1)10

⇒10n = 100

⇒ n= 100/10

⇒ n= 10

10+20+…+100= (10/2)[2(10)+(10-1)(10)]

= (10/2)(20+90)

Therefore, the required sum is:

= 2550+ 1050 – 550

Hence, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

Question 8 : Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, 0 < x < 10}; B = {2, 3, 5, 7}. Write the set (A U B)’.

Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, 0 < x < 10}; B = {2, 3, 5, 7}

U = { 1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {2, 4, 6, 8}

A U B = {2, 3, 4, 5, 6, 7, 8}

(A U B)’ = {1, 9}

Find More: Differences between Pure Maths and Applied Maths

Question 9 : In a survey of 600 students in a school, 150 students were drinking Tea and 225 drinking Coffee, and 100 were drinking both Tea and Coffee. Find how many students were drinking neither Tea nor Coffee.

Total number of students = 600

Number of students who were drinking Tea = n(T) = 150

Number of students who were drinking Coffee = n(C) = 225

Number of students who were drinking both Tea and Coffee = n(T ∩ C) = 100

n(T U C) = n(T) + n(C) – n(T ∩ C)

= 150 + 225 -100

= 375 – 100

Hence, the number of students who are drinking neither Tea nor Coffee = 600 – 275 = 325

Find More: CBSE Class 11 Commerce Syllabus

## NCERT Solution for Class 11 Applied Math Algebra

NCERT Solutions for CBSE Class 11 Applied Mathematics Algebra is an essential resource for preparing for the exam. You can verify your answers and understand how each problem is solved.

NCERT Applied Maths Algebra solutions provide a straightforward way of illustrating and explanations. The format of these textbooks is straightforward.

## Complex Number Class 11 Notes with Examples

Complex Numbers Class 11 is defined when a number can be represented in the form p + iq. Here, p and q are real numbers and i =−1−−−√. For a complex number z = p + iq, p is known as the real part, represented by Re z, and q is known as the imaginary part, represented by Im z of complex number z.

The topics and the subtopics taught in Complex Numbers CBSE class 11 applied maths algebra are:

- Introduction
- Complex Numbers
- Algebra of Complex Numbers - Addition of two complex Numbers; The difference between two complex Numbers; Multiplication of two complex Numbers; Division of two complex Numbers; Power of i; The square root of a negative real number; Identities
- The Modulus and the Conjugate of Complex Numbers
- Argand Plane and Polar Representation - Polar Representation of Complex Numbers

The notes for class 11 give you detailed knowledge describing the concepts involved in complex numbers. Some of the examples are:

Example 1: If 4x + i(3x – y) = 3 + i (– 6), where x and y are real numbers, then find x and y values.

Solution: Given,

4x + i (3x – y) = 3 + i (–6) ….(1)

By equating the real and the imaginary parts of equation (1),

4x = 3, 3x – y = –6,

Now, 4x = 3

⇒ x = 3/4

And 3x – y = -6

⇒ y = 3x + 6

Substituting the value of x,

⇒ y = 3(3/4) + 6

⇒ y = 33/4

Therefore, x = 3/4 and y = 33/4.

Example 2 : Express (-√3 + √-2)(2√3 – i) in the form of a + ib.

Solution: We know that i2 = -1

(-√3 + √-2)(2√3 – i) = (-√3 + i√2)(2√3 – i)

= (-√3)(2√3) + (i√3) + i(√2)(2√3) – i2√2

= -6 + i√3(1 + 2√2) + √2

= (-6 + √2) + i√3(1 + 2√2)

This is of the form a + ib, where a = -6 + √2 and b = √3(1 + √2).

Example 3: Find the multiplicative inverse of 2 – 3i.

Solution: Let z = 2 – 3i

z ¯ = 2 + 3i

|z|2 = (2)2 + (-3)2 = + = 13

We know that the multiplicative inverse of z is given by the formula:

z −1= z ¯| z |2

= (2 + 3i)/13

= (2/13) + i(3/13)

Alternatively,

Multiplicative inverse of z is:

z-1 = 1/(2 – 3i)

By rationalizing the denominator we get,

= (2 + 3i)/(4 + 9)

= (2 + 3i)/ 13

Example 4: Represent the complex number z = 1 + i√3 in the polar form.

Solution: Given, z = 1 + i√3

Let 1 = r cos θ, √3 = r sin θ

By squaring and adding, we get

r2(cos2θ + sin2θ) = 4

r = 2 (as r > 0)

Therefore, cos θ = 1/2 and sin θ = √3/2

This is possible when θ = π/3.

Thus, the required polar form is z = 2[cos π/3 + i sin π/3].

Hence, the complex number z = 1 + i√3 is represented as shown in the below figure.

In this blog, we delved into the world of CBSE Class 11 Applied Maths Coordinate Geometry for 2024. We explored concepts, project topics, syllabus, weightage, and crucial preparation tips.

Key Takeaways:

- CBSE's commitment to making mathematics more applied and relatable with the introduction of Class 11 Applied Mathematics.
- The substantial weightage of algebra in the syllabus, commanding 15 marks in the theory exam.
- The significance of mastering algebraic fundamentals.
- Common mistakes students should avoid while solving algebra problems.
- A glimpse of solved sample questions to enhance your understanding.

Find More :CBSE Class 11 Humanities Subjects List

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Frequently Asked Questions

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Case Study Questions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities. Last modified on: 9 months ago; Reading Time: 1 Minute; ... Algebraic Expressions and Identities; Maths: CBSE Class 8: Chapter Covered: Class 8 Maths Chapter 9: Topics: Type of Questions: Case Study Questions:

Important class 8 maths questions for chapter 9 Algebraic expressions and Identities will help students to get better prepared for CBSE class 8 exam and develop problem-solving skills. These algebraic expressions and identities questions not only cover NCERT questions but also other variations of questions to help class 8 students get ...

Multiplying Monomial by Monomial. This lesson covers skills from the following lessons of the NCERT Math Textbook: (i) 8.2 Multiplication of Algebraic Expressions (ii) 8.3 Multiplying a Monomial by Monomial.

He has designed the window, using square and, rectangular pieces of glass, as shown below., , , , Use the algebraic identities to solve the questions., , 1,, , If one side of the pink square measures 14 inches and one side of the grey square measures, 6 inches, write the area of the window as the sum of three numbers., , a. 14+844+6 b. 14+ 168 ...

Ans: Given: \ [x\] We need to find the coefficient of the given expression. We know that coefficient is the number before any variable. So, the expression \ [x\] can be written as $ 1\times x $ . So, the coefficient will be $ 1. $. 2. The expression for sum of numbers p and q subtracted from their product is.

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BYJU'S NCERT Solutions for Class 8 Maths provide detailed solutions to all questions covered in Chapter 9 Algebraic Expressions and Identities. Our experienced academicians provide detailed and step-by-step solutions for all the questions in the subject of Maths from the NCERT textbook. Download the free Class 8 Maths NCERT Solutions and get ...

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Algebraic Expressions and Identities Class 8 Notes is specially designed to help students understand important chapter concepts clearly and study productively. These notes will also enable students to have an effective math practice session and be ready to tackle chapter questions that could be asked in the exams.

Updated fornew NCERT.Get NCERT Solutions ofChapter 8 Class 8 Algebraic Expressions and Identitiesfree at Teachoo. Answers to all exercise questions, examples have been solved with step-by-step solutions. Concepts are explained before doing the questions.In this chapter, we will learnWhat arealgebra.

Class 8 Maths Chapter 8 | Algebraic Expressions & Identities | Case Study QuestionIn this video, I have solved case study question of class 8 maths chapter 8...

The NCERT Solutions Class 8 Maths Chapter 9 covers the concept of Algebraic Expressions and Identities in detail with the help of proper explanations and most importantly by introducing the components which make up an expression that is the 'terms'. It further zeroes on what makes up a term" that is the 'coefficient' and 'variable ...

Question 8. Multiply the following expressions: (a) 3xy 2 × (-5x 2 y) (b) 1 2 x 2 yz × 2 3 xy 2 z × 1 5 x 2 yz. Solution: Question 9. Find the area of the rectangle whose length and breadths are 3x 2 y m and 5xy 2 m respectively. Solution: Length = 3x 2 y m, breadth = 5xy 2 m.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5. Ex 9.5 Class 8 Maths Question 1. Ex 9.5 Class 8 Maths Question 2. Use the identity (x + a) (x + b) = x 2 + (a + b)x + ab to find the following products. Ex 9.5 Class 8 Maths Question 3. Find the following squares by using the identities.

Get Revision notes of Class 8th Mathematics Chapter 9 Algebraic expressions and identities to score good marks in your Exams. Our notes of Chapter 9 Algebraic expressions and identities are prepared by Maths experts in an easy to remember format, covering all syllabus of CBSE, KVPY, NTSE, Olympiads, NCERT & other Competitive Exams.

Below is a sample list of questions and their solutions from our Class 8 Mathematics Chapter 9 Important Questions: Question 1. Find the terms and their coefficients for each of the following expressions. (i) 5xyz² - 3zy. (ii) 1 + x+ x². (iii) 4x²y² - 4x²y²z² + z². (iv) 3- pq + qr - rp. (v) x/2 + y/2 - xy.

Ans: Class 8 Maths introduces the concept of algebraic expressions and algebraic identities to the students. Algebraic identities are algebraic equations that are valid for all variable values in them. Following are the formulas for algebraic expressions: (a+b)2 = a2+2ab +b2. (a-b)2= a2-2ab +b2.

Multiple choice questions (MCQs) are available for Class 8 chapter 9 Algebraic Expressions and Identities chapter. For all the problems, there are four multiple options given among which one is the right answer. Solve each question and choose the correct answer. 1. An algebraic expression that contains only one term is called: A. Monomial. B ...

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. To score good marks in Class 8 Mathematics examination, it is advised to solve questions provided at the end of each chapter in the ...

Chapter 9 Class 8 Maths NCERT Ex 9.1 Solutions. The first exercise of Algebraic Expressions and Identities includes 4 questions. This exercise is based on the introduction to algebraic expressions, terms, coefficients, monomials, binomials, and polynomial expressions. It also includes questions related to like and unlike terms and addition and ...

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NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.1. Ex 9.1 Class 8 Maths Question 1. Identify the terms, their coefficients for each of the following expressions. (i) 5xyz 2 - 3zy. (ii) 1 + x + x 2. (iii) 4x 2 y 2 - 4x 2 y 2 z 2 + z 2. (iv) 3 - pq + qr - rp.

In this blog, we delved into the world of CBSE Class 11 Applied Maths Coordinate Geometry for 2024. We explored concepts, project topics, syllabus, weightage, and crucial preparation tips. CBSE's commitment to making mathematics more applied and relatable with the introduction of Class 11 Applied Mathematics.