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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 of algebraic expressions.

• 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 Monomial by Monomial

• 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)
• Multiply monomials by polynomials Get 3 of 4 questions to level up!
• 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!

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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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Revision Notes on Algebraic Expressions and Identities

Algebraic expression.

Any mathematical expression which consists of numbers, variables and operations are called Algebraic Expression .

Every expression is separated by an operation which is called Terms . Like 7n and 2 are the two terms in the above figure.

Every term is formed by the product of the factors .7n is the product of 7 and n which are the factors of 7n.

3. Coefficient

The number placed before the variable or the numerical factor of the term is called Coefficient of that variable .7 is the numerical factor of 7n so 7 is coefficient here.

4. Variable

Any letter like x, y etc. are called Variables. The variable in the above figure is n.

5. Operations

Addition, subtraction etc. are the operations which separate each term.

6. Constant

The number without any variable is constant . 2 is constant here.

Number Line and an Expression

An expression can be represented on the number line.

How to represent x + 5 and x – 5 on the number line?

First, mark the distance x and then x + 5 will be 5 unit to the right of x.

In the case of x – 5 we will start from the right and move towards the negative side. x – 5 will be 5 units to the left of x.

Monomials, Binomials and Polynomials

Like and Unlike Terms

Terms having the same variable are called Like Terms .

Examples of Like Terms

24xy and 5yx

6x 2 and 12x 2

The terms having different variable are called, Unlike Terms .

Examples of Unlike Terms

2x and - 9y

24xy and 5pq

6x 2 and 12y 2

Addition and Subtraction of Algebraic Expressions

Steps to add or Subtract Algebraic Expression

First of all, we have to write the algebraic expressions in different rows in such a way that the like terms come in the same column.

Add them as we add other numbers.

If any term of the same variable is not there in another expression then write is as it is in the solution.

Add 15p 2 – 4p + 5 and 9p – 11

Write down the expressions in separate rows with like terms in the same column and add. 

Subtract 5a 2 – 4b 2 + 6b – 3 from 7a 2 – 4ab + 8b 2 + 5a – 3b.

For subtraction also write the expressions in different rows. But to subtract we have to change their signs from negative to positive and vice versa.

Multiplication of Algebraic Expressions

While multiplying we need to take care of some points about the multiplication of like and unlike terms.

1. Multiplication of Like Terms

The coefficients will get multiplied.

The power will not get multiplied but the resultant variable will be the addition of the individual powers.

The product of 4x and 3x will be 12x 2 .

The product of 5x, 3x and 4x will be 60x 3 .

2. Multiplication of Unlike Terms

The power will remain the same if the variable is different.

If some of the variables are the same then their powers will be added.

The product of 2p and 3q will be 6pq

The product of 2x 2 y, 3x and 9 will  be 54x 3 y

Multiplying a Monomial by a Monomial

1. Multiplying Two Monomials

While multiplying two polynomials the resultant variable will come by

The coefficient of product = Coefficient of the first monomial × Coefficient of the second monomial

The algebraic factor of product = Algebraic factor of the first monomial × Algebraic factor of the second monomial.

25y × 3xy = 125xy 2

2. Multiplying Three or More Monomials

While multiplying three or more monomial the criterion will remain the same.

4xy × 5x 2 y 2 × 6x 3 y 3 = (4xy × 5x 2 y 2 ) × 6x 3 y 3

= 20x 3 y 3 × 6x 3 y 3

= 120x 3 y 3 × x 3 y 3

= 120 (x 3 × x 3 ) × (y 3 × y 3 )

= 120x 6 × y 6

= 120x 6 y 6

We can do it in other way also

4xy × 5x 2 y 2 × 6x 3 y 3

= (4 × 5 × 6) × (x × x 2 × x 3 ) × (y × y 2 × y 3 )

= 120 x 6 y 6

Multiplying a Monomial by a Polynomial

1. Multiplying a Monomial by a Binomial

To multiply a monomial with a binomial we have to multiply the monomial with each term of the binomial.

Multiplication of 8 and (x + y) will be (8x + 8y).

Multiplication of 3x and (4y + 7) will be (12xy + 21x).

Multiplication of 7x 3  and (2x 4  + y 4 ) will be (14x 7 + 7x 3 y 4 ).

2. Multiplication of Monomial by a trinomial

This is also the same as above.

Multiplication of 8 and (x + y + z) will be (8x + 8y + 8z).

Multiplication of 4x and (2x + y + z) will be (8x 2  + 4xy + 4xz).

Multiplication of 7x 3  and (2x 4 + y 4 + 2) will be (14x 7  + 7x 3 y 4  + 14x 3 ).

Multiplying a Polynomial by a Polynomial

1. Multiplying a Binomial by a Binomial

We use the distributive law of multiplication in this case. Multiply each term of a binomial with every term of another binomial. After multiplying the polynomials we have to look for the like terms and combine them.

Simplify (3a + 4b) × (2a + 3b)

(3a + 4b) × (2a + 3b)

= 3a × (2a + 3b) + 4b × (2a + 3b)    [distributive law]

= (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b)

= 6 a 2 + 9ab + 8ba + 12b 2

= 6 a 2 + 17ab + 12b 2      [Since ba = ab]

2. Multiplying a Binomial by a Trinomial

In this also we have to multiply each term of the binomial with every term of trinomial.

Simplify (p + q) (2p – 3q + r) – (2p – 3q) r.

 We have a binomial (p + q) and one trinomial (2p – 3q + r)

(p + q) (2p – 3q + r)

= p(2p – 3q + r) + q (2p – 3q + r)

= 2p 2 – 3pq + pr + 2pq – 3q 2 + qr

= 2p 2 – pq – 3q 2 + qr + pr    (–3pq and 2pq are like terms)

(2p – 3q) r = 2pr – 3qr

(p + q) (2p – 3q + r) – (2p – 3q) r

= 2p 2 – pq – 3q 2 + qr + pr – (2pr – 3qr)

= 2p 2 – pq – 3q 2 + qr + pr – 2pr + 3qr

= 2p 2 – pq – 3q 2 + (qr + 3qr) + (pr – 2pr)

= 2p 2 – 3q 2 – pq + 4qr – pr

An identity is an equality which is true for every value of the variable but an equation is true for only some of the values of the variables.

So an equation is not an identity.

Like, x 2 = 1, is valid if x is 1 but is not true if x is 2.so it is an equation but not an identity.

Some of the Standard Identities

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

(a - b) 2 = a 2 – 2ab + b 2

a 2 – b 2 = (a + b) (a - b)

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

(a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca

These identities are useful in carrying out squares and products of algebraic expressions. They give alternative methods to calculate products of numbers and so on.

Applying Identities

(4x – 3y) 2

= (4x) 2 – 2(4x) (3y) + (3y) 2 

= 16x 2 – 24xy + 9y 2

Use the Identity (x + a) (x + b) = x 2 + (a + b) x + ab to find the value of 501 × 502

 501 × 502

= (500 + 1) × (500 + 2)

 = 500 2 + (1 + 2) × 500 + 1 × 2

= 250000 + 1500 + 2

= 251502  

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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.

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

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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.

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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.

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

Algebraic Expressions and Identities Class 8 Questions And Answers provided here. These NCERT Solutions are prepared by experts team at Careers360 team considering the latest syllabus and pattern of CBSE 2023-24. Th is chapter will introduce you to the application of algebraic terms and variables to solve various problems. Algebra is the most important branch of mathematics which teaches how to form equations and solving them using different kinds of techniques . Important topics like the product of the equation, finding the coefficient of the variable in the equation , subtraction of the equation and creating quadratic equation by the product of its two roots, and division of the equation are covered in this chapter. Also Practice NCERT solutions for class 8 maths to command the concepts.

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Algebraic Expressions and Identities Class 8 Questions And Answers PDF Free Download

Algebraic expressions and identities class 8 solutions - important formulae, algebraic expressions and identities class 8 ncert solutions (intext questions and exercise), ncert solutions for class 8 maths chapter 9 algebraic expressions and identities - topics, ncert solutions for class 8 maths - chapter wise, ncert solutions for class 8 - subject wise, also check ncert books and ncert syllabus here.

Download PDF

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

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

Free download NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities for CBSE Exam.

NCERT Solutions to Exercises of Chapter 9: Algebraic Expressions and Identities

what are expressions?

Question: 1 Give five examples of expressions containing one variable and five examples of expressions containing two variables.

Five examples of expressions containing one variable are:

Five examples of expressions containing two variables are:

Question: 2(i) Show on the number line

x on the number line:

Question: 2(ii) Show on the number line :

x-4 on the number line:

Question: 2(iii) Show on the number line :

2x+1 on the number line:

Question: 2(iv) Show on the number line:

3x - 2 on the number line

Algebraic expressions and identities class 8 solutions - Topic 9.2 Terms, Factors and Coefficients

Question:1 Identify the coefficient of each term in the expression.

coefficient of each term are given below

Algebraic expressions and identities class 8 ncert solutions - Topic 9.3 Monomials, Binomials and Polynomials

Question: 1(i) Classify the following polynomials as monomials, binomials, trinomials.

Binomial since there are two terms with non zero coefficients.

Question: 1(ii) Classify the following polynomials as monomials, binomials, trinomials.

Trinomial since there are three terms with non zero coefficients.

Question:1(iii) Classify the following polynomials as monomials, binomials, trinomials.

Question: 1(iv) Classify the following polynomials as monomials, binomials, trinomials.

Question: 1(v) Classify the following polynomials as monomials, binomials, trinomials.

Monomial since there is only one term.

Three binomials with the only x as a variable are:

Three binomials with x and y as variables are:

Three monomials with x and y as variables are

Question: 2(d) Construct 2 polynomials with 4 or more terms .

Two polynomials with 4 or more terms are:

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Topic 9.4 Like and Unlike Terms

Question:(i) Write two terms which are like

Question:(ii) Write two terms which are like

we can write more like terms

Question:(iii) Write two terms which are like

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities - Exercise: 9.1

Question:1(i) Identify the terms, their coefficients for each of the following expressions.

following are the terms and coefficient

Question: 1(ii) Identify the terms, their coefficients for each of the following expressions.

the following is the solution

Question:1(iii) Identify the terms, their coefficients for each of the following expressions.

Question: 1(iv) Identify the terms, their coefficients for each of the following expressions.

The terms are 3, -pq, qr,and -rp and the coefficients are 3, -1, 1 and -1 respectively.

Question:1(v) Identify the terms, their coefficients for each of the following expressions.

Above are the terms and coefficients

Question: 1(vi) Identify the terms, their coefficients for each of the following expressions.

The terms are 0.3a, -0.6ab and 0.5b and the coefficients are 0.3, -0.6 and 0.5.

Question: 2(a) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Question: 2(b) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Question: 2(c) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

This polynomial does not fit in any of these three categories.

Question: 2(d) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Question: 2(e) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Question: 2(f) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Question: 2(g) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Question: 2(h) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Question: 2(i) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Question:2(j) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Question: 2(k) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Question: 3(i) Add the following.

ab-bc+bc-ca+ca-ab=0.

Question:3 (ii) Add the following.

Question:3 (iii) Add the following

Question: 3(iv) Add the following.

12a-9ab+5b-3-(4a-7ab+3b+12) =(12-4)a +(-9+7)ab+(5-3)b +(-3-12) =8a-2ab+2b-15

NCERT class 8 maths chapter 9 question answer - Topic 9.7.2 Multiplying Three or More Monomials

We observe that the result is same in both cases and the result does not depend on the order in which multiplication has been carried out.

Class 8 maths chapter 9 question answer - exercise: 9.2

Question: 1(i) Find the product of the following pairs of monomials.

Question: 1(ii) Find the product of the following pairs of monomials.

Question: 1(iii) Find the product of the following pairs of monomials

Question: 1(iv) Find the product of the following pairs of monomials.

Question:1(v) Find the product of the following pairs of monomials.

Question:2(A) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

The question can be solved as follows

Question:2(B) Find the areas of rectangles with the following pairs of monomials as their lengths and breadth respectively.

the area is calculated as follows

Question:2(C) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

Question:2(D) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

area of rectangles is

Question:2(E) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

The area is calculated as follows

Question:3 Complete the table of products.

Question:4(i) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

Question:4(ii) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

the volume of rectangular boxes with the following length, breadth and height is

Question:4(iii) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

Question:4(iv) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

Question:5(i) Obtain the product of

the product

Question:5(ii) Obtain the product of

Question:5(iii) Obtain the product of

Question:5(iv) Obtain the product of

Question:5(v) Obtain the product of

Class 8 maths chapter 9 NCERT solutions - Topic 9.8.1 Multiplying a Monomial by a Binomial

Question:(i) Find the product

Using distributive law,

Question:(ii) Find the product

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities - Topic 9.8.2 Multiplying A Monomial By A Trinomial

Question:1 Find the product:

By using distributive law,

Class 8 maths chapter 9 NCERT solutions - exercise: 9.3

Question:1(i) Carry out the multiplication of the expressions in each of the following pairs.

Multiplication of the given expression gives :

By distributive law,

Question:1(ii) Carry out the multiplication of the expressions in each of the following pairs.

We have ab, (a-b).

Using distributive law we get,

Question:1(iii) Carry out the multiplication of the expressions in each of the following pairs.

Using distributive law we can obtain multiplication of given expression:

Question:1(iv) Carry out the multiplication of the expressions in each of the following pairs.

We will obtain multiplication of given expression by using distributive law :

Question:1(v) Carry out the multiplication of the expressions in each of the following pairs.

Using distributive law :

Question:2 Complete the table

We will use distributive law to find product in each case.

Question:3(i) Find the product.

Opening brackets :

Question:3(ii) Find the product.

Question:3(iii) Find the product.

Question:3(iv) Find the product.

(a) We have

. So We get,

Put a = 1 ,

Put a = (-1)

(a)First we will solve each brackets individually.

Firstly, open the brackets:

Adding both, we get :

At first we will solve each bracket individually,

Subtracting:

Solving brackets :

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities-Exercise: 9.4

Question:1(i) Multiply the binomials.

Question:1(ii) Multiply the binomials.

Question:1(iii) Multiply the binomials

Question:1(iv) Multiply the binomials.

(a + 3b) X (x + 5) = (a)(x) + (a)(5) + (3b)(x) + (3b)(5) = ax + 5a + 3bx + 15b

Question:1(v) Multiply the binomials.

(2pq + 3q 2 ) X (3pq - 2q 2 ) = (2pq)(3pq) + (2pq)(-2q 2 ) + ( 3q 2 )(3pq) + (3q 2 )(-2q 2 ) = 6p 2 q 2 - 4pq 3 + 9pq 3 - 6q 4 = 6p 2 q 2 +5pq 3 - 6q 4

Question:1(vi) Multiply the binomials.

Multiplication can be done as follows

Question:2(i) Find the product.

Question:2(ii) Find the product.

Question:2(iii) Find the product.

Question:2(iv) Find the product.

following is the solution

Question:3(i) Simplify.

this can be simplified as follows

Question:3(ii) Simplify .

This can be simplified as

Question:3(iii) Simplify.

simplifications can be

Question:3(iv) Simplify.

(a + b) X ( c -d) + (a - b) X (c + d) + 2(ac + bd ) = (a)(c) + (a)(-d) + (b)(c) + (b)(-d) + (a)(c) + (a)(d) + (-b)(c) + (-b)(d) + 2(ac + bd ) = ac - ad + bc - bd + ac +ad -bc - bd + 2(ac + bd ) = 2(ac - bd ) + 2(ac +bd ) = 2ac - 2bd + 2ac + 2bd = 4ac

Question:3(v) Simplify.

Question:3(vi) Simplify.

simplification is done as follows

Question:3(vii) Simplify.

Question:3(viii) Simplify.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities - Topic 9.11 Standard Identities

Question:1(i) Put -b in place of b in identity 1. Do you get identity 2?

NCERT Free Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities - Topic 9.11 Standard Identities

Question:2 Consider, the special case of Identity (IV) with a = b, what do you get? Is it related to Identity

Class 8 algebraic expressions and identities NCERT solutions - exercise: 9.5

Question:1(i) Use a suitable identity to get each of the following products.

Question:1(ii) Use a suitable identity to get each of the following products in bracket.

Question:1(iii) Use a suitable identity to get each of the following products in bracket.

Question:1(iv) Use a suitable identity to get each of the following products in bracket.

Question:1(v) Use a suitable identity to get each of the following products in bracket.

Question:1(vi) Use a suitable identity to get each of the following products in bracket.

Question:1(vii) Use a suitable identity to get each of the following.

Question:1(viii) Use a suitable identity to get each of the following product.

Question:1(ix) Use a suitable identity to get each of the following product.

Question:1(x) Use a suitable identity to get each of the following products.

Question:3(i) Find the following squares by using the identities.

Question:3(ii) Find the following squares by using the identities.

Question:3(iii) Find the following squares by using the identities.

Question:3(iv) Find the following squares by using the identities.

Question:3(v) Find the following squares by using the identities.

Question:3(vi) Find the following squares by using the identities.

Question:4(i) Simplify:

Question:4(ii) Simplify.

remember that

here a= 2x, b= 5

Question:4(iii) Simplify.

Question: 4(iv) Simplify.

Question: 4(v) Simplify.

Question:4(vi) Simplify.

Question:4(vii) Simplify.

Question:5(i) Show that

Hence it is prooved

Question:5(ii) Show that

Question:5(iii) Show that.

First we will solve the LHS :

Question:5(iv) Show that.

Opening both brackets we get,

Question:5(v) Show that

Opening all brackets from the LHS, we get :

Question:6(i) Using identities, evaluate.

We will use the identity:

Question:6(ii) Using identities, evaluate.

Here we will use the identity :

Question:6(iii) Using identities, evaluate.

Question:6(iv) Using identities, evaluate.

Here we will the identity :

Question:6(v) Using identities, evaluate.

Here we will use :

Question:6(vi) Using identities, evaluate.

This can be written as :

Question:6(vii) Using identities, evaluate.

This can be written in form of :

Question:6(viii) Using identities, evaluate.

Question:6(ix) Using identities, evaluate.

Using this formula,

= (2.00)(0.04)

Here x =100, a = 3, b = 4

Here x =5, a = 0.1, b = 0.2

Here x =100, a = 3, b = -2

Here x =10, a = -0.3, b = -0.2

• What are Expressions?
• Terms, Factors and Coefficients
• Monomials, Binomials,f and Polynomials
• Like and Unlike Terms
• Addition and Subtraction of Algebraic Expressions
• Multiplication of Algebraic Expressions: Introduction
• Multiplying a Monomial by a Monomial
• Multiplying two monomials
• Multiplying three or more monomials
• Multiplying a Monomial by a Polynomial
• Multiplying a monomial by a binomial
• Multiplying a monomial by a trinomial
• Multiplying a Polynomial by a Polynomial
• Multiplying a binomial by a binomial
• Multiplying a binomial by a trinomial
• What is an Identity?
• Standard Identities
• Applying Identities
• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science

Some Important Identities From NCERT Book for Class 8 Chapter 9 Algebraic Expressions And Identities

you can write

Can be simplified as follows

Now add each term

You can form the above identities by yourself. These above identities have been used in many problems of NCERT solutions for class 8 maths chapter 9 algebraic expression and identities.

• NCERT Books Class 8 Maths
• NCERT Syllabus Class 8 Maths
• NCERT Books Class 8
• NCERT Syllabus Class 8

Frequently Asked Question (FAQs)

Addition and subtraction of the algebraic expression, multiplication of the algebraic expression, standard identities, and application of identities are important topics in this chapter.

CBSE class 8 maths is not tough at all. It teaches a very basic and simple maths.

There are 16 chapters starting from rational number to playing with numbers in the CBSE class 8 maths.

Here you will get the detailed  NCERT solutions for class 8  by clicking on the link.

Here you will get the detailed  NCERT solutions for class 8 maths  by clicking on the link.

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

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Data Administrator

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

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The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary. 

Ethical Hacker

A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

GIS officer work on various GIS software to conduct a study and gather spatial and non-spatial information. GIS experts update the GIS data and maintain it. The databases include aerial or satellite imagery, latitudinal and longitudinal coordinates, and manually digitized images of maps. In a career as GIS expert, one is responsible for creating online and mobile maps.

Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

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Database Architect

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For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

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Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism , Advertising , Marketing Management . Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

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For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication) , B.M.M. (Bachelor of Mass Media) , or  MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

Production Manager

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

ITSM Manager

Automation test engineer.

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process. 

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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| pdf download.

• Exercise 9.1 Chapter 9 Class 8 Maths NCERT Solutions
• Exercise 9.2 Chapter 9 Class 8 Maths NCERT Solutions
• Exercise 9.3 Chapter 9 Class 8 Maths NCERT Solutions
• Exercise 9.4 Chapter 9 Class 8 Maths NCERT Solutions
• Exercise 9.5 Chapter 9 Class 8 Maths NCERT Solutions

NCERT Solutions for Class 8 Maths Chapters:

How many exercises in chapter 9 algebraic expressions and identities, identify the variable and constant in the expression 24 – x., the length and breadth of a rectangle are 3x 2 – 2 and 2x + 5 respectively. find its area., verify the identity (x + a)( x + b) = x 2 + (a + b)x + ab for a = 2, b = 3 and x = 4,, contact form.

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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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Algebraic Expressions and Identities Class 8 MCQ Test (Online Available)

Free mcq test, table of content, algebraic expressions and identities test - 39.

Duration: 10 Mins

Maximum Marks: 10

Read the following instructions carefully.

1. The test contains 10 total questions.

2. Each question has 4 options out of which only one is correct .

3. You have to finish the test in 10 minutes.

4. You will be awarded 1 mark for each correct answer.

5. You can view your Score & Rank after submitting the test.

6. Check detailed Solution with explanation after submitting the test.

7. Rank is calculated on the basis of Marks Scored & Time

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The chapter Algebraic Expressions and Identities is one of the important chapters in class 8. The highly qualified experts of Selfstudys developed these Algebraic Expressions and Identities class 8 MCQ to test what students have learnt and also helps them to identify their strengths and weaknesses. 

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• Chapter 9: Algebric Expressions And Identities
• Exercise 9.1

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.1

The NCERT solutions for Class 8 CBSE include topics with frequent, focused, engaging challenges and activities that strengthen Math concepts. Each question of exercise 9.1 in NCERT Class 8 Maths Solutions has been carefully solved and helps you obtain the correct answer.

Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.1 questions and answers help students to understand polynomial terms, factors and coefficients as well as addition and subtraction of the polynomials. NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.1 are prepared by BYJU’S subject experts using a step-by-step approach. Download free NCERT Solutions for Maths Chapter 9 and practise offline.

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Access Answers of Maths NCERT Class 8 Chapter 9 Algebraic Expressions and Identities Exercise 9.1 Page number 140

Exercise 9.1 page no: 140.

• 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 – p (v) (x/2) + (y/2) – xy (vi) 0.3a – 0.6ab + 0.5b

2. Classify the following polynomials as monomials, binomials, trinomials. 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 – 3y 2 , 2y – 3y 2  + 4y 3 , 5x – 4y + 3xy, 4z – 15z 2 , ab + bc + cd + da, pqr, p 2 q + pq 2 , 2p + 2q

Let us first define the classifications of these 3 polynomials:

Monomials, contain only one term.

Binomials, contain only two terms.

Trinomials, contain only three terms.

3.  Add the following. (i) ab – bc, bc – ca, ca – ab (ii) a – b + ab, b – c + bc, c – a + ac (iii) 2p 2 q 2  – 3pq + 4, 5 + 7pq – 3p 2 q 2 (iv) l 2  + m 2 , m 2  + n 2 , n 2  + l 2 , 2lm + 2mn + 2nl

i) (ab – bc) + (bc – ca) + (ca-ab) = ab – bc + bc – ca + ca – ab = ab – ab – bc + bc – ca + ca

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

= a – a +b – b +c – c + ab + bc + ca

=0 + 0 + 0 + ab + bc + ca

= ab + bc + ca

iii) 2p 2 q 2  – 3pq + 4, 5 + 7pq – 3p 2 q 2 = (2p 2 q 2  – 3pq + 4) + (5 + 7pq – 3p 2 q 2 ) = 2p 2 q 2  – 3p 2 q 2  – 3pq + 7pq + 4 + 5 = – p 2 q 2  + 4pq + 9

iv)  (l 2  + m 2 ) + (m 2  + n 2 ) + (n 2  + l 2 ) + (2lm + 2mn + 2nl) = l 2  + l 2  + m 2  + m 2  + n 2  + n 2  + 2lm + 2mn + 2nl = 2l 2  + 2m 2  + 2n 2  + 2lm + 2mn + 2nl

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

(b) Subtract  3xy + 5yz – 7zx  from  5xy – 2yz – 2zx + 10xyz

 (c) Subtract  4p 2 q – 3pq + 5pq 2  – 8p + 7q – 10 from  18 – 3p – 11q + 5pq – 2pq 2  + 5p 2 q

(a) (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

b) (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

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

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

=18+10 -3p+8p -11q – 7q + 5 pq+ 3pq- 2pq^2 – 5pq^2 + 5 p^2 q – 4p^2 q

= 28 + 5p – 18q + 8pq – 7pq 2  + p 2 q

Access Other Exercise Solutions of Class 8 Maths Chapter 9 Algebraic Expressions and Identities

Exercise 9.2 Solutions : 5 Questions (Short answers)

Exercise 9.3 Solutions : 5 Questions (Short answers)

Exercise 9.4 Solutions : 3 Questions (Short answers)

Exercise 9.5 Solutions : 8 Questions (Short answers)

This exercise of  Class 8 Maths Chapter 9 Algebraic Expressions and Identities is based on the polynomial terms, factors and coefficients, and addition and subtraction of algebraic expressions. The operations like addition and subtraction are mainly performed on monomials, binomials and polynomials. After practising these questions, students will become confident about the concept and able to solve problems on their own.

Also, explore – 

NCERT Solutions for Class 8 

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• Class 8 NCERT Solutions - Chapter 9 Algebraic Expressions and Identities - Exercise 9.2
• Class 8 NCERT Solutions - Chapter 9 Algebraic Expressions and Identities - Exercise 9.1
• Class 8 NCERT Solutions - Chapter 9 Algebraic Expressions and Identities - Exercise 9.3
• Class 8 NCERT Solutions- Chapter 9 Algebraic Expressions and Identities - Exercise 9.4
• Class 8 NCERT Solutions - Chapter 9 Algebraic Expressions and Identities - Exercise 9.5 | Set 2
• Class 8 NCERT Solutions - Chapter 9 Algebraic Expressions and Identities - Exercise 9.5 | Set 1
• Class 8 RD Sharma Solutions- Chapter 6 Algebraic Expressions And Identities - Exercise 6.2
• Class 8 RD Sharma Solutions - Chapter 6 Algebraic Expressions And Identities - Exercise 6.1
• Class 8 RD Sharma Solutions - Chapter 6 Algebraic Expressions and Identities - Exercise 6.3 | Set 2
• Class 8 RD Sharma Solutions - Chapter 6 Algebraic Expressions And Identities - Exercise 6.6 | Set 2
• Class 8 RD Sharma Solutions - Chapter 6 Algebraic Expressions And Identities - Exercise 6.7
• Class 8 RD Sharma Solutions- Chapter 6 Algebraic Expressions And Identities - Exercise 6.6 | Set 1
• Class 8 RD Sharma Solutions - Chapter 6 Algebraic Expressions And Identities - Exercise 6.3 | Set 1
• Class 9 RD Sharma Solutions - Chapter 4 Algebraic Identities- Exercise 4.2
• Class 8 NCERT Solutions- Chapter 12 Exponents and Powers - Exercise 12.1
• Class 11 NCERT Solutions- Chapter 8 Binomial Theorem - Exercise 8.1
• Class 9 NCERT Solutions - Chapter 2 Polynomials - Exercise 2.5 | Set 2
• Class 8 RD Sharma Solutions - Chapter 2 Powers - Exercise 2.1
• Class 9 NCERT Solutions - Chapter 2 Polynomials - Exercise 2.5 | Set 1

NCERT Solutions Class 8 – Chapter 8 Algebraic Expressions and Identities – Exercise 8.2

Question 1. find the product of the following pairs of monomials..

Monomial: Expression containing only one term

(i) 4, 7p 

Ans:  

 (4) * (7p) = 28p

 (ii) -4p, 7p

Ans:   

(-4p) * (7p)  = -28p 2 Explanation: When a negative number is multiplied to a positive number the product becomes negative.

(iii) -4p, 7pq

(-4p) * (7pq)  = -28p 2 q

(iv) 4p 3 , -3p

(4p 3 ) * (-3p) = -12p 4

Ans:    

(4p) * (0) = 0  Explanation: Any number when multiplied to zero (0) gives zero.

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

           (p, q); (10m, 5n); (20x 2 , 5y 2 ); (4x, 3x 2 ); (3mn, 4np).

Note: Area of rectangle is the product of length and breadth [length * breadth]

          For (p,q):                        p * q = pq           For (10m, 5n):                       10m *  5n = 50mn          For (20x 2 ,5y 2 ):                        20x 2 * 5y 2 = 100x 2 y 2           For (4x,3x 2 ):                          4x * 3x 2 = 12x 3           For (3mn, 4np):                            3mn * 4np = 12mn 2 p

Question 3. Complete the table of products.

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

Note: The volume of the rectangle is the product of length, breadth, height [length * breadth * height]

(i) 5a, 3a 2 , 7a 4

 5a * 3a 2 * 7a 4 = 105a 7

(ii) 2p , 4q , 8r

2p * 4q * 8r = 64pqr

(iii) xy, 2x 2 y, 2xy 2

xy * 2x 2 y * 2xy 2 = 4x 4 y 4

(iv) a, 2b, 3c

Ans:  a * 2b * 3c = 6abc

Question 5. Obtain the product of 

(i) xy, yz, zx

xy * yz * zx = x 2 y 2 z 2

(ii) a, -a 2 , a 3

a * -a 2 * a 3 = -a 6

(iii) 2, 4y, 8y 2 , 16y 3

2 * 4y * 8y 2 * 16y 3 = 1024y 6

(iv) a, 2b, 3c, 6abc

a * 2b * 3c * 6abc = 36a 2 b 2 c 2

(v) m, -mn, mnp

m * -mn * mnp = -m 3 n 2 p

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