polynomials case study questions class 10

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CBSE Class 10 Maths Case Study Questions for Chapter 2 - Polynomials (Published by CBSE)

Check the case study questions published by cbse for class 10 maths chapter 2 - polynomials. these questions are important for the preparation of cbse class 10 maths exam 2021-22..

CBSE Class 10 Maths paper in Board Exam 2022 will have some questions based on the case study. These questions are entirely new for the class 10 students. Therefore, the board has released a question bank to help the students get familiarised with the case study questions. We have provided here the case study questions for CBSE Class 10 Maths Chapter 2 - Polynomials. All the questions have sub-questions of MCQ type. You can find the answer (correct option) written against each question. Practice all the case study based questions right after you finish with the chapter - Polynomials. This will help you prepare for your Maths exam easily and effectively.

Case Study Questions for Class 10 Maths Chapter 2 - Polynomials

CASE STUDY 1:

The below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.

1. In the standard form of quadratic polynomial, ax 2 + bx + c, a, b and c are

a) All are Polynomials.

b) All are rational numbers.

c) ‘a’ is a non zero real number and b and c are any Polynomials.

d) All are integers.

Answers: c) ‘a’ is a non zero real number and b and c are any Polynomials.

2. If the roots of the quadratic polynomial are equal, where the discriminant D = b 2 – 4ac, then

a) D > 0

b) D < 0

c) D ≥ 0

Answers: d) D = 0

3. If α and 1/α are the zeroes of the quadratic polynomial 2x2 – x + 8k, then k is

c) –1/4

Answers: b) 1/4

4. The graph of x 2 +1 = 0

a) Intersects x‐axis at two distinct points.

b)Touches x‐axis at a point.

c) Neither touches nor intersects x‐axis.

d)Either touches or intersects x‐ axis.

Answers: c) Neither touches nor intersects x‐axis.

5. If the sum of the roots is –p and product of the roots is –1/p, then the quadratic polynomial is

a) k(–px 2 + x/p + 1)

b) k(px 2 – x/p – 1)

c) k(x 2 + px – 1/p)

d) k(x 2 – px + 1/p)

Answers: c) k(x 2 + px – 1/p)

CASE STUDY 2:

An asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.

1. The shape of the poses shown is

d) Parabola

Answer: d) Parabola

2. The graph of parabola opens downwards, if _______

a) a ≥ 0

c) a < 0

d) a > 0

Answer: c) a < 0

3. In the graph, how many zeroes are there for the polynomial?

Answer: c) 2

4. The two zeroes in the above shown graph are

Answer: b) -2, 4

CASE STUDY 3:

Basketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoor on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing quadratic polynomial.

1. The shape of the path traced shown is

2. The graph of parabola opens upwards, if _______

b) a < 0

c) a > 0

d) a ≥ 0

Answer: c) a > 0

3. Observe the following graph and answer

In the above graph, how many zeroes are there for the polynomial?

Answer: d) 3

4. The three zeroes in the above shown graph are

b) -2, 3, 1

c) -3, -1, 2

d) -2, -3, -1

Answer: c) -3, -1, 2

5. What will be the expression of the polynomial?

a) x 3 + 2x 2 – 5x – 6

b) x 3 + 2x 2 – 5x + 6

c) x 3 + 2x 2 + 5x – 6

d) x 3 + 2x 2 + 5x + 6

Answer: a) x 3 + 2x 2 – 5x – 6

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

Tips to Solve Case Study Based Questions Accurately

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CBSE 10th Standard Maths Subject Polynomials Case Study Questions With Solution 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

Case Study Questions

(ii) The expression of the polynomial represented by the graph is

(iii) Find the value of the polynomial represented by the graph when x = 6.

(iv) The sum of zeroes of the polynomial x 2  + 2x - 3 is

(v) If the sum of zeroes of polynomial at 2  + 5t + 3a is equal to their product, then find the value of a.

(ii) Find the value of \(\alpha\) + \(\beta\) +  \(\alpha\) \(\beta\) .

(iii) The value of p(2) is

(iv) If \(\alpha\) and \(\beta\) are zeroes of  \(x^{2}+x-2, \text { then } \frac{1}{\alpha}+\frac{1}{\beta}=\)

(v) If sum of zeroes of  \(q(x)=k x^{2}+2 x+3 k\)  is equal to their product, then k =

(ii) The axis of symmetry of the given parabola is

(iii) The zeroes of the polynomial, represented in the given graph, are

(iv) Which of the following polynomial has -2 and -3 as its zeroes?

(v) For what value of 'x', the value of the polynomial  \(f(x)=(x-3)^{2}+9 \text { is } 9 ?\)

(ii) What will be the expression of the polynomial given in diagram?

(iii) What is the value of the polynomial, represented by the graph, when x = 4?

(iv) If the tunnel is represented by x 2  + 3x - 2, then its zeroes are

(v) If one zero is 4 and sum of zeroes is -3, then representation of tunnel as a polynomial is

(ii) The sum of product of zeroes taken two at a time is

(iii) Product of zeroes of polynomial p(x) is

(iv) The value of the polynomial p(x), when x = 4 is

(v) If  \(\alpha,\beta,\gamma\)  are the zeroes of a polynomial g(x) such that  \(\alpha+\beta+\gamma=3, \alpha \beta+\beta \gamma+\gamma \alpha=-16\)   and  \(\alpha \beta \gamma=-48\)   then, g(x) =

*****************************************

Cbse 10th standard maths subject polynomials case study questions with solution 2021 answer keys.

(i) (b): Graph of a quadratic polynomial is a parabolic in shape. (ii) (c): Since the graph of the polynomial cuts the x-axis at (-6,0) and (6, 0). So, the zeroes of polynomial are -6 and 6. \(\therefore\) Required polynomial is p(x) = x 2 - (-6 + 6)x + (-6)(6) = x 2 - 36 (iii) (c) : We have, p(x) = x 2 - 36 Now, p( 6) = 62 - 36 = 36 - 36 = 0 (iv) (b): Letf (x) = x 2 + 2x - 3. Then, \(\text { Sum of zeroes }=-\frac{\text { coefficient of } x}{\text { coefficient of } x^{2}}=-\frac{(2)}{1}=-2\) (v) (d): The given polynomial is at 2 + 5t + 3a Given, sum of zeroes = product of zeroes. \(\Rightarrow \quad \frac{-5}{a}=\frac{3 a}{a} \Rightarrow a=\frac{-5}{3}\)

(i) (b): Given, a and \(\beta\) are the zeroes of  \(p(x)=x^{2}-24 x+128\) \(\text { Putting } p(x)=0 \text { , we get }\) \( x^{2}-8 x-16 x+128=0 \) \(\Rightarrow x(x-8)-16(x-8)=0 \) \(\Rightarrow (x-8)(x-16)=0 \Rightarrow x=8 \text { or } x=16 \) \(\therefore \alpha=8, \beta=16\) (ii) (c) :  \(\alpha+\beta+\alpha \beta =8+16+(8)(16) =24+128=152 \) (iii) (d) :  \(p(2)=2^{2}-2 4(2)+128=4-48+128=84\) (iv) (a): Since a and \(\beta\) are zeroes of  \(x^{2}+x-2\) \(\therefore \quad \alpha+\beta=-1 \text { and } \alpha \beta=-2 \) \(\text { Now, } \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}=\frac{-1}{-2}=\frac{1}{2}\) (v) (c): Sum of zeroes  \(=\frac{-2}{k}\) Product of zeroes  \(=\frac{3 k}{k}=3\) According to question, we have  \(\frac{-2}{k}=3\) \(\Rightarrow \quad k=\frac{-2}{3}\)

(i) (b): The shape of the path of the soccer ball is a parabola. (ii) (c): The axis of symmetry of the given curve is a line parallel to y-axis. (iii) (a): The zeroes of the polynomial, represented in the given graph, are -2 and 7, since the curve cuts the x-axis at these points. (iv) (d): A polynomial having zeroes -2 and -3 is  \(p(x)=x^{2}-(-2-3) x+(-2)(-3)=x^{2}+5 x+6\) (v) (c): We have  \(f(x)=(x-3)^{2}+9\) \(\text { Now, } 9=(x-3)^{2}+9 \) \(\Rightarrow(x-3)^{2}=0 \Rightarrow x-3=0 \Rightarrow x=3\)

(i) (a): Since, the graph intersects the x-axis at two points, namely x = 8, -2. So, 8, - 2 are the zeroes of the given polynomial. (ii) (b): The expression of the polynomial given in diagram is  \(-x^{2}+6 x+16\) (iii) (c) : Let  \(p(x)=-x^{2}+6 x+16\) \(\text { When } x=4, p(4)=-4^{2}+6 \times 4+16=24\) (iv) (d): Let  \(f(x)=-x^{2}+3 x-2\) Now, consider  \(f(x)=0 \Rightarrow-x^{2}+3 x-2=0\) \(\begin{aligned} &\Rightarrow x^{2}-3 x+2=0 \Rightarrow(x-2)(x-1)=0\\ &\Rightarrow x=1,2 \text { are its zeroes. } \end{aligned}\) (v) (b): Let a and \(\beta\)  are the zeroes of the required polynomial. Given  \(\alpha + \beta = - 3\) If \(\alpha\) = 4, then  \(\beta\) = -7 \(\therefore \quad \text { Representation of tunnel is }-x^{2}-3 x+28 \text { . }\)

(i) (c): For finding  \(\alpha,\beta,\)   \(\gamma\)  consider p(x) = 0 \(\Rightarrow \quad x^{3}-18 x^{2}+95 x-150=0 \) \(\Rightarrow \quad(x-3)\left(x^{2}-15 x+50\right)=0 \) \(\Rightarrow \quad(x-3)(x-5)(x-10)=0 \Rightarrow x=10 \text { or } x=5 \text { or } x=3 \) \(\text { Thus } \alpha=10, \beta=5 \text { and } \gamma=3\) (ii) (d): Here  \(\alpha=10, \beta=5 \text { and } \gamma=3\) \(\therefore\)   Sum of product of zeroes taken two at a time \(\begin{array}{l} =\alpha \beta+\beta \gamma+\gamma \alpha=(10)(5)+(5)(3)+(3)(10) \\ =50+15+30=95 \end{array}\) (iii) (a): Product of zeroes of polynomial p(x) =  \(\alpha\beta\gamma\) = (10) (5) (3) = 150 (iv) (b): We have  \(p(x)=x^{3}-18 x^{2}+95 x-150\) \(\begin{array}{l} \text { Now, } p(4)=4^{3}-18(4)^{2}+95(4)-150 \\ =64-288+380-150=6 \end{array}\) (v) (d):   \(g(x)=x^{3}-(\alpha+\beta+\gamma) x^{2} +(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma \) \(\Rightarrow g(x)=x^{3}-3 x^{2}-16 x-(-48)=x^{3}-3 x^{2}-16 x+48\)

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Class 10 Maths Case Study Questions Chapter 2 Polynomials

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Case study Questions in the Class 10 Mathematics Chapter 2  are very important to solve for your exam. Class 10 Maths Chapter 2 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Class 10 Maths Case Study Questions Chapter 2  Polynomials

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In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Polynomials Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 2 Polynomials

Case Study/Passage-Based Questions

Question 1:

Basketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoors on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of a parabola representing quadratic polynomial.

1. The shape of the path traced shown is

d) Parabola

Answer: d) Parabola

2. The graph of parabola opens upwards, if _______

b) a < 0

c) a > 0

Answer: c) a > 0

3. Observe the following graph and answer

In the above graph, how many zeroes are there for the polynomial?

Answer: d) 3

4. The three zeroes in the above shown graph are

b) -2, 3, 1

c) -3, -1, 2

d) -2, -3, -1

Answer: c) -3, -1, 2

5. What will be the expression of the polynomial?

a) x 3  + 2x 2  – 5x – 6

b) x 3  + 2x 2  – 5x + 6

c) x 3  + 2x 2  + 5x – 6

d) x 3  + 2x 2  + 5x + 6

Answer: a) x3 + 2×2 – 5x – 6

Answer: (b) parabolic

(ii) The expression of the polynomial represented by the graph is

Answer: (c) x2-36

(iii) Find the value of the polynomial represented by the graph when x = 6.

Answer: (c) 0

(iv) The sum of zeroes of the polynomial x 2  + 2x – 3 is

Answer: (b) -2

(v) If the sum of zeroes of polynomial at 2  + 5t + 3a is equal to their product, then find the value of a.

Answer:  (d) −53

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 2 Polynomials with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Polynomials Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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CBSE Case Study Questions for Class 10 Maths Polynomials Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Polynomials in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 10 Maths Polynomials PDF

Mcq set 1 -, mcq set 2 -, checkout our case study questions for other chapters.

• Chapter 1: Real Numbers Case Study Questions
• Chapter 3: Pair of Linear Equations in Two Variables Case Study Questions
• Chapter 4: Quadratic Equation Case Study Questions
• Chapter 5: Arithmetic Progressions Case Study Questions

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Class 10 Maths Chapter 2 MCQ

Class 10 Maths Chapter 2 MCQ based on Case Study with answers and explanation for the first term exams 2024-25. All the questions including CBSE MCQ set of questions are answered with complete explanation. Class 10 Maths Chapter 2 Case study based MCQ of Polynomials helps the students to understand the pattern of CBSE question papers and way of answering.

During the skipping through skipping rope, its look like the in the form of parabola. It is a natural examples of parabolic shape which is represented by a quadratic polynomial. Similarly, we can observe in many other cases forming a in a variety of forms of different parabolas.

In the standard form of quadratic polynomial, ax² + bx + c, the condition between a, b and c are

Because if a = 0, the equation become linear. Then it is not a quadratic polynomial. Hence, the correct option is (C).

• View Answer

If the roots of the quadratic polynomial are unequal, where the discriminant D = b² – 4ac, then

We know that: If D = b² – 4ac 0, real and unequal roots. Hence, the correct option is (A).

If α and -α are the zeroes of the quadratic polynomial 2x² – 3(k – 4)x – 8, then k is

From the quadratic polynomial 2x² – 3(k – 4)x – 8 Sum of zeros = 3(k – 4)/2 [Because sum of zeros = -b/a] So, α + (- α) = 3(k – 4)/2, therefore, k – 4 = 0 or k = 4 Hence, the correct option is (A).

The graph of x² – 1 = 0

The given equation: x² – 1 = 0 or 1.x² + 0.x – 1 = 0 Here, a = 1, b = 0 and c = – 1 D = b² – 4ac = 0² – 4 х 1 х(- 1) = 4, therefore, there is two real root. So, the graph of the equation intersects x‐axis at two distinct points. Hence, the correct option is (A).

If the sum of the roots is p and product of the roots is -p, then the quadratic polynomial is

We know that the equation of a quadrilateral is given by: k [x² – (sum of roots)x + product of roots)] Therefore, the equation = k(x² – (p)x + (-p)] = k(x² – px – p) Hence, the correct option is (C).

The below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.

In the standard form of quadratic polynomial, ax² + bx + c, a, b and c are

Because if a = 0, the equation become linear. Then it is not a quadratic equation. Hence, the correct option is (C).

If the roots of the quadratic polynomial are equal, where the discriminant D = b² – 4ac, then

We know that: If D = b² – 4ac 0, real and unequal roots. Hence, the correct option is (D).

If α and 1/ α are the zeroes of the quadratic polynomial 2x² – x + 8k, then k is

From the quadratic polynomial 2x² – x + 8k Product of zeros = 8k/4 [Because product of zeros = c/a] So, α x (1/α) = 8k/2 Therefore, 4k = 1 Hence, k = 1/4 Hence, the correct option is (B).

The graph of x² + 1 = 0

The given equation: x² + 1 = 0 or 1.x² + 0.x + 1 = 0 Here, a = 1, b = 0 and c = 1 D = b² – 4ac = 0² – 4 х 1 х 1 = – 4, therefore, there is no real root. So, the graph of the equation neither touches nor intersects x‐axis. Hence, the correct option is (C).

If the sum of the roots is –p and product of the roots is -1/p, then the quadratic polynomial is

We know that the equation of a quadrilateral is given by: k [x² – (sum of roots)x + product of roots)] Therefore, the equation = k(x² – (-p)x + (-1/p)] = k(x² + px – 1/p) Hence, the correct option is (C).

Observe the position of the athlete taking long jump. He use to follow every time a particular shape of path. In the figure, a student can observe that the different positions can be related to representation of quadratic polynomial.

The path of the different positions form a

When we draw a dotted line following the different positions of athlete, it show a parabolic path. So, it is a parabola. Hence, the correct option is (D).

If the above case is represented by quadratic polynomial ax² + bx + c, then

If a > 0, the parabola is upward. If a < 0, the parabola is downward. Hence, the correct option is (C).

If the sum of zeros of quadratic polynomial ax² + bx + c is equal to product of zero, then

Given polynomial: ax² + bx + c Sum of zeros = -b/a, Product of zeros = c/a, According to question, -b/a = c/a Therefore, – b = a or a + b = 0 Hence, the correct option is (C).

Observe the graph given below.

In the above graph, how many zeroes are there for the polynomial?

The graph of the polynomial cutting the x-axis at four distinct points. So, it has 4 zeros. Hence, the correct option is (D).

The four zeroes in the above shown graph are

The graph of the polynomial cutting x-axis at (-4, 0), (-2, 0), (1, 0) and (3, 0). So, the zeros of polynomial are -4, -2, 1 and 3. Hence, the correct option is (A).

An asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.

The shape of the poses shown is

The pose shown above are in the format of parabola. Hence, the correct option is (D).

The graph of parabola opens downwards, if

The zeroes of the quadratic polynomial 4√3 x² + 5x – 2√3 are.

Given polynomial: 4√3 x² + 5x – 2√3 = 4√3 x² + 8x – 3x – 2√3 = 4x (√3x + 2) – √3 (√3x + 2) = (√3x + 2) (4x – √3) Putting (√3x + 2) (4x – √3) = 0 x = -2/√3 or x = √3/4 Hence, the correct option is (B).

Observe the following graph:

In the graph, how many zeroes are there for the polynomial?

The graph of the polynomial cutting the x-axis at two distinct points. So, it has two zeros. Hence, the correct option is (C).

The two zeroes in the above shown graph are

The graph of the polynomial cutting x-axis at (-2, 0) and (4, 0). So, the zeros of polynomial are -2 and 4. Hence, the correct option is (B).

Basketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoor on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing quadratic polynomial.

The shape of the path traced shown is

The graph of parabola opens upwards, if.

Observe the following graph and answer:

The graph of the polynomial cutting the x-axis at three distinct points. So, it has three zeros. Hence, the correct option is (D).

The three zeroes in the above shown graph are

The graph of the polynomial cutting x-axis at (-3, 0), (-1, 0) and (2, 0). So, the zeros of polynomial are -3, -1 and 2. Hence, the correct option is (C).

What will be the expression of the polynomial?

The zeros of polynomial are -3, -1 and 2. Let α = -3, β = -1 and γ = 2 So, α + β + γ = -3 + (-1) + 2 = – 4 + 2 = -2 αβ + βγ + αγ = (-3)(-1) + (-1)(2) + (2)(-3) = 3 – 2 – 6 = – 5 αβγ = (-3)(-1)(2) = 6 Equation of polynomial = x³ – (α + β + γ) x² + (αβ + βγ + αγ) x – αβγ = x³ – (-2)x² + (- 5)x – 6 = x³ + 2x² – 5x – 6

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Case Study Based Questions for Class 10 Maths Chapter 2 Polynomials

• Last modified on: 8 months ago
• Reading Time: 4 Minutes

Case Study Questions for Class 10 Maths Chapter 1 Real Numbers

Question 1:

Priya and her husband Aman who is an architect by profession, visited France. They went to see Mont Blanc Tunnel which is a highway tunnel between France and Italy, under the Mont Blanc Mountain in the Alps, and has a parabolic cross-section. The mathematical representation of the tunnel is shown in the graph.

Based on the above information, answer the following questions.

(1) The zeroes of the polynomial whose graph is given, are (a) -2, 8 (b)-2, -8 (c) 2,8 (d) -2,0

(2) What will be the expression of the polynomial given in diagram? (a) x 2 – 6x + 16 (b) -x 2 + 6x + 16 (c) x 2 + 6x + 16 (d) -x 2 – 6x – 16

(3) What is the value of the polynomial. represented by the graph, when x = 4? (a) 22 (b) 23 (c) 24 (d) 25

(4) If the tunnel is represented by -x 2 +3x-2. then its zeroes are (a) -1, -2 (b) 1, -2 (c) -1, 2 (d) 1, 2

(5) If one of the zero is 4 and sum of zeroes is -3, then representation of tunnel as a polynomial is (a) x 2 -x+24 (b)-x 2 -3x+28 (c) x 2 +x+28 (d) x 2 -x+28

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CBSE Class 10 Maths Case Study

CBSE Board has introduced the case study questions for the ongoing academic session 2021-22. The board will ask the paper on the basis of a different exam pattern which has been introduced this year where 50% syllabus is occupied for MCQ for Term 1 exam. Selfstudys has provided below the chapter-wise questions for CBSE Class 10 Maths. Students must solve these case study based problems as soon as they are done with their syllabus. 

These case studies are in the form of Multiple Choice Questions where students need to answer them as asked in the exam. The MCQs are not that difficult but having a deep and thorough understanding of NCERT Maths textbooks are required to answer these. Furthermore, we have provided the PDF File of CBSE Class 10 maths case study 2021-2022.

Class 10 Maths (Formula, Case Based, MCQ, Assertion Reason Question with Solutions)

In order to score good marks in the term 1 exam students must be aware of the Important formulas, Case Based Questions, MCQ and Assertion Reasons with solutions. Solving these types of questions is important because the board will ask them in the Term 1 exam as per the changed exam pattern of CBSE Class 10th.

Important formulas should be necessarily learned by the students because the case studies are solved with the help of important formulas. Apart from that there are assertion reason based questions that are important too. 

Assertion Reasoning is a kind of question in which one statement (Assertion) is given and its reason is given (Explanation of statement). Students need to decide whether both the statement and reason are correct or not. If both are correct then they have to decide whether the given reason supports the statement or not. In such ways, assertion reasoning questions are being solved. However, for doing so and getting rid of confusions while solving. Students are advised to practice these as much as possible.

For doing so we have given the PDF that has a bunch of MCQs questions based on case based, assertion, important formulas, etc. All the Multiple Choice problems are given with detailed explanations.

CBSE Class 10th Case study Questions

Recently CBSE Board has the exam pattern and included case study questions to make the final paper a little easier. However, Many students are nervous after hearing about the case based questions. They should not be nervous because case study are easy and given in the board papers to ease the Class 10th board exam papers. However to answer them a thorough understanding of the basic concepts are important. For which students can refer to the NCERT textbook.

Basically, case study are the types of questions which are developed from the given data. In these types of problems, a paragraph or passage is given followed by the 5 questions that are given to answer . These types of problems are generally easy to answer because the data are given in the passage and students have to just analyse and find those data to answer the questions.

CBSE Class 10th Assertion Reasoning Questions

These types of questions are solved by reading the statement, and given reason. Sometimes these types of problems can make students confused. To understand the assertion and reason, students need to know that there will be one statement that is known as assertion and another one will be the reason, which is supposed to be the reason for the given statement. However, it is students duty to determine whether the statement and reason are correct or not. If both are correct then it becomes important to check, does reason support the statement? 

Moreover, to solve the problem they need to look at the given options and then answer them.

CBSE Class 10 Maths Case Based MCQ

CBSE Class 10 Maths Case Based MCQ are either Multiple Choice Questions or assertion reasons. To solve such types of problems it is ideal to use elimination methods. Doing so will save time and answering the questions will be much easier. Students preparing for the board exams should definitely solve these types of problems on a daily basis.

Also, the CBSE Class 10 Maths MCQ Based Questions are provided to us to download in PDF file format. All are developed as per the latest syllabus of CBSE Class Xth.

Class 10th Mathematics Multiple Choice Questions

Class 10 Mathematics Multiple Choice Questions for all the chapters helps students to quickly revise their learnings, and complete their syllabus multiple times. MCQs are in the form of objective types of questions whose 4 different options are given and one of them is a true answer to that problem. Such types of problems also aid in self assessment.

Case Study Based Questions of class 10th Maths are in the form of passage. In these types of questions the paragraphs are given and students need to find out the given data from the paragraph to answer the questions. The problems are generally in Multiple Choice Questions.

The Best Class 10 Maths Case Study Questions are available on Selfstudys.com. Click here to download for free.

To solve Class 10 Maths Case Studies Questions you need to read the passage and questions very carefully. Once you are done with reading you can begin to solve the questions one by one. While solving the problems you have to look at the data and clues mentioned in the passage.

In Class 10 Mathematics the assertion and reasoning questions are a kind of Multiple Choice Questions where a statement is given and a reason is given for that individual statement. Now, to answer the questions you need to verify the statement (assertion) and reason too. If both are true then the last step is to see whether the given reason support=rts the statement or not.

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CBSE Class 10 Maths: Case Study Questions of Chapter 2 Polynomials PDF Download

Case study Questions in the Class 10 Mathematics Chapter 2  are very important to solve for your exam. Class 10 Maths Chapter 2 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based   questions for Class 10 Maths Chapter 2  Polynomials

In CBSE Class 10 Maths Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Polynomials Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 2 Polynomials

Case Study/Passage-Based Questions

Question 1:

Basketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoors on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of a parabola representing quadratic polynomial.

1. The shape of the path traced shown is

d) Parabola

Answer: d) Parabola

2. The graph of parabola opens upwards, if _______

b) a < 0

c) a > 0

Answer: c) a > 0

3. Observe the following graph and answer

In the above graph, how many zeroes are there for the polynomial?

Answer: d) 3

4. The three zeroes in the above shown graph are

b) -2, 3, 1

c) -3, -1, 2

d) -2, -3, -1

Answer: c) -3, -1, 2

5. What will be the expression of the polynomial?

a) x 3  + 2x 2  – 5x – 6

b) x 3  + 2x 2  – 5x + 6

c) x 3  + 2x 2  + 5x – 6

d) x 3  + 2x 2  + 5x + 6

Answer: a) x3 + 2×2 – 5x – 6

Answer: (b) parabolic

(ii) The expression of the polynomial represented by the graph is

Answer: (c) x2-36

(iii) Find the value of the polynomial represented by the graph when x = 6.

Answer: (c) 0

(iv) The sum of zeroes of the polynomial x 2  + 2x – 3 is

Answer: (b) -2

(v) If the sum of zeroes of polynomial at 2  + 5t + 3a is equal to their product, then find the value of a.

Answer:  (d) −53

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 2 Polynomials with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Polynomials Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Chapter 2 Class 10 Polynomials

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Updated the chapter according to the new NCERT Book for 2023-24 session

Get NCERT Solutions of C hapter 2 Class 10 Polynomials free at Teachoo. All NCERT Exercise Questions, Examples and Optional Questions have been solved with video of each and every question.

In this chapter, we will learn

• What is a polynomial
• What are monomial, binomials, trinomials
• What is the degree of polynomial
• What are linear, quadratic and cubic polynomials
• What are Zeroes of a Polynomials

• How to find Zeroes of a Polynomials (both quadratic and cubic)
• Quadratic Polynomial
• Cubic Polynomial
• Dividing two polynomials, and verifying the Division Algorithm for Polynomials

We have divided this chapter into 2 parts - Serial Order Wise and Concept Wise.

In Serial Order Wise, there are examples and exercises just like the NCERT Book. It is useful when we are looking to get an answer of a particular question.

That is not a good way of doing the chapter.

The best way to learning maths is through Concept Wise 

Concept wise is Teachoo's (टीचू) way of learning.

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10th Class Mathematics Polynomials Question Bank

Done case based (mcqs) - polynomials total questions - 40.

A) all are real numbers done clear

B) all are rational numbers done clear

C) 'a' is a non-zero real number and b and c are any real numbers done clear

D) all are integers done clear

question_answer 2) If the roots of the quadratic polynomial are equal, and the discriminant \[D={{b}^{2}}-4ac,\] then:

A) \[D>0\] done clear

B) \[D<0\]      done clear

C) \[D\ge 0\] done clear

D) \[D=0\] done clear

question_answer 3) If \[\alpha \] and \[\frac{1}{\alpha }\]are the zeroes of the quadratic polynomial \[2{{x}^{2}}-x+8k,\] then k is:

A) \[4\] done clear

B) \[\frac{1}{4}\] done clear

C) \[\frac{-1}{4}\] done clear

D) \[2\] done clear

question_answer 4) The graph of \[{{\text{x}}^{\text{2}}}+\text{4}=0:\]

A) intersects X-axis at two distinct points done clear

B) touches X-axis at a point done clear

C) neither touches nor intersects X-axis done clear

D) either touches or intersects X-axis. done clear

question_answer 5) If the sum of the roots is \[-p\] and product of the roots \[-\frac{1}{p}\] is, then the quadratic polynomial is:

A) \[k\left( -p{{x}^{2}}+\frac{x}{p}+1 \right)\] done clear

B) \[k\left( p{{x}^{2}}-\frac{x}{p}-1 \right)\] done clear

C) \[k\left( {{x}^{2}}+px-\frac{1}{p} \right)\] done clear

D) \[k\left( {{x}^{2}}-px+\frac{1}{p} \right)\] done clear

A) spiral done clear

B) ellipse done clear

C) linear done clear

D) parabola done clear

question_answer 7) The graph of parabola opens downwards, if..........

A) \[a\ge 0\] done clear

B) \[a=0\] done clear

C) \[a<0\] done clear

D) \[a>0\] done clear

A) 0 done clear

B) 1 done clear

C) 2 done clear

D) 3 done clear

question_answer 9) The quadratic polynomial of the two zeroes in the above shown graph are:

A) \[k({{x}^{2}}-2x-8)\] done clear

B) \[k({{x}^{2}}+2x-8)\] done clear

C) \[k({{x}^{2}}+2x+8)\] done clear

D) \[k({{x}^{2}}-2x+8)\] done clear

question_answer 10) The   zeroes   of   the   quadratic   polynomial \[4\sqrt{3}\,{{x}^{2}}+5x-2\sqrt{3}\] are:

A) \[\frac{2}{\sqrt{3}}.\frac{\sqrt{3}}{4}\] done clear

B) \[-\frac{2}{\sqrt{3}}.\frac{\sqrt{3}}{4}\] done clear

C) \[\frac{2}{\sqrt{3}}.-\frac{\sqrt{3}}{4}\] done clear

D) \[-\frac{2}{\sqrt{3}}.-\frac{\sqrt{3}}{4}\] done clear

A) \[(2,-4)\] done clear

B) \[(4,-2)\] done clear

C) \[(-2,-2)\] done clear

D) \[(-4,-4)\] done clear

A) Intersects X-axis done clear

B) Intersects V-axis done clear

C) Intersects /-axis or X-axis done clear

D) None of these done clear

question_answer 13) Graph of a quadratic polynomial is a:

A) straight line done clear

B) circle done clear

C) parabola done clear

D) ellipse done clear

question_answer 14) The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is:

A) \[{{x}^{2}}-6x+2\] done clear

B) \[{{x}^{2}}-36\] done clear

C) \[{{x}^{2}}-6\] done clear

D) \[{{x}^{2}}-3\] done clear

question_answer 15) The number of zeroes that polynomial \[f(x)={{(x-2)}^{2}}+4\] can have is:

A) 1 done clear

B) 2 done clear

C) 0 done clear

A) Parabola done clear

B) Hyperbola done clear

C) Ellipse done clear

question_answer 17) The zeroes of given quadratic polynomial are:

A) \[2,-4\] done clear

B) \[-2,4\] done clear

C) \[3,-4\] done clear

D) \[-3,4\] done clear

question_answer 18) Read from the graph the value of y corresponding to \[x=-1\] is:

A) \[-8\] done clear

B) \[-6\] done clear

C) \[-5\] done clear

D) \[-2\] done clear

question_answer 19) The graph of the given quadratic polynomial cut at which points on the X-axis?

A) \[(-2,0),\,\,(4,0)\] done clear

B) \[(0,-2),\,\,(0,4)\] done clear

C) \[(0,-2),\,\,(0,-8)\] done clear

D) \[None\,\, of\,\, the\,\,above\] done clear

question_answer 20) The graph of the given quadratic polynomial cut at which point on X-axis?

A) \[(-8,0)\] done clear

B) \[(0,-8)\] done clear

C) \[(-10,0)\] done clear

D) \[(-10,0)\] done clear

C) 3 done clear

D) 4 done clear

question_answer 22) Which type of polynomial is represented by Jay's graph?

A) Linear done clear

B) Parabola done clear

C) Zig-zag done clear

question_answer 23) How many zeroes are there for the Richa's graph?

question_answer 24) If \[p(x)=a{{x}^{2}}+bx+c\] and \[a+b+c=0,\] then one zero is:

A) \[\frac{-b}{a}\]  done clear

B) \[\frac{c}{a}\] done clear

C) \[\frac{b}{c}\] done clear

D) \[-\frac{c}{a}\] done clear

question_answer 25) If \[p(x)=a{{x}^{2}}+bx+c\] and \[a+c=b,\] then one of the zeroes is:

A) \[\frac{b}{a}\] done clear

C) \[\frac{-c}{a}\] done clear

D) \[\frac{-b}{a}\] done clear

A) Yes done clear

B) No done clear

C) Can't say done clear

question_answer 27) If yes, then the correct quadratic polynomial is:

A) \[4\sqrt{3}{{x}^{2}}-5x+2\sqrt{3}\] done clear

B)        \[4\sqrt{3}{{x}^{2}}+5x-2\sqrt{3}\] done clear

C) \[4\sqrt{3}{{x}^{2}}+5x+2\sqrt{3}\] done clear

D) \[4\sqrt{3}{{x}^{2}}-5x-2\sqrt{3}\] done clear

question_answer 28) The value of \[{{\alpha }^{2}}+{{\beta }^{2}}\] is:

A) \[\frac{53}{48}\] done clear

B) \[\frac{59}{48}\] done clear

C) \[\frac{73}{48}\] done clear

D) \[\frac{71}{48}\] done clear

question_answer 29) What is the value of the correct polynomial if \[x=-1\]?

A) \[-5+2\sqrt{3}\] done clear

B) \[5-2\sqrt{3}\] done clear

C) \[5-6\sqrt{3}\] done clear

D) \[-5+6\sqrt{3}\] done clear

question_answer 30) The value of \[{{\alpha }^{3}}-{{\beta }^{3}}\]is:

A) \[\frac{-539}{192\sqrt{3}}\] done clear

B) \[\frac{539}{192\sqrt{3}}\] done clear

C) \[\frac{539\sqrt{3}}{192}\] done clear

D) \[\frac{-539\sqrt{3}}{192}\] done clear

A) \[\frac{3}{5}\] done clear

B) \[\frac{3}{4}\] done clear

C) \[\frac{2}{5}\] done clear

D) \[\frac{1}{5}\] done clear

question_answer 32) The zeroes of the quadratic polynomial \[{{x}^{2}}+20x+96\]are:

A) both positive done clear

B) both negative done clear

C) one positive and other negative done clear

D) both equal done clear

question_answer 33) The quadratic polynomial whose zeroes are 5 and \[-12\] is given by:

A) \[{{x}^{2}}+7x-60\] done clear

B) \[15{{x}^{2}}-x-6\] done clear

C) \[{{x}^{2}}-7x+60\] done clear

D) \[15{{x}^{2}}+x+6\] done clear

question_answer 34) If one zero of the polynomial \[f(x)=5{{x}^{2}}+13x+m\] is reciprocal of the other, then the value of m is:

A) 6 done clear

B) 0 done clear

C) 5 done clear

D) 1 done clear

question_answer 35) Which of the following cannot be the graph, of a quadratic polynomial?

question_answer 37) If the product of the zeroes of the quadratic polynomial \[f(x)=a{{x}^{2}}-6x-6\] is 4, then the value of a is:

A) \[\frac{-3}{2}\] done clear

B) \[\frac{3}{2}\] done clear

C) \[\frac{2}{3}\] done clear

D) \[\frac{-2}{3}\] done clear

question_answer 38) The flow of the water in the pool is represented by \[{{x}^{2}}-2x-8,\] then its zeroes are:

B) \[4,-2\] done clear

C) \[2,-2\] done clear

D) \[-4,-4\] done clear

question_answer 39) If a and p be the zeroes of the polynomial \[{{x}^{2}}-1,\]then the value of \[\frac{1}{\alpha }+\frac{1}{\beta }\]is:

A) \[0\] done clear

B) \[\frac{1}{2}\] done clear

C) \[1\] done clear

D) \[-1\] done clear

question_answer 40) A quadratic polynomial whose one zero is \[-3\] and product of zeroes is 0, is:

A) \[3{{x}^{2}}+3\] done clear

B) \[{{x}^{2}}-3x\] done clear

C) \[{{x}^{2}}+3x\] done clear

D) \[3{{x}^{2}}-3\] done clear

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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Ncert solutions class 10 maths chapter 2 – cbse free pdf download.

NCERT Solutions Class 10 Maths Chapter 2 Polynomials are provided here to help the students in learning efficiently for their exams. The subject experts of Maths have prepared these solutions to help students prepare well for their board exams. They have solved these solutions in such a way that it becomes easier for students to practise the questions of Chapter 2 Polynomials using the Solutions of NCERT . This makes it simple for the students to learn by adding step-wise explanations to these Maths NCERT Class 10 Solutions.

Download Exclusively Curated Chapter Notes for Class 10 Maths Chapter – 2 Polynomials

Download most important questions for class 10 maths chapter – 2 polynomials.

NCERT Solutions for Class 10 Maths is an extremely important study resource for students. Solving these Polynomials NCERT Solutions of Class 10 Maths would help the students fetch good marks in the board exams. Moreover, experts have focused on following the updated CBSE Syllabus for 2023-24 while preparing these solutions.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability
• Exercise 2.1 Chapter 2 Polynomials
• Exercise 2.2 Chapter 2 Polynomials
• Exercise 2.3 Chapter 2 Polynomials
• Exercise 2.4 Chapter 2 Polynomials

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Exercise 2.1 page: 28.

1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Graphical method to find zeroes:-

Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis.

(i) In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does not cut it at any point.

(ii) In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at only one point.

(iii) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at any three points.

(iv) In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at two points.

(v) In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at four points.

(vi) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at three points.

Exercise 2.2 Page: 33

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x 2 –2x –8

⇒ x 2 – 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)

Therefore, zeroes of polynomial equation x 2 –2x–8 are (4, -2)

Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x 2 )

Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x 2 )

(ii) 4s 2 –4s+1

⇒4s 2 –2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)

Therefore, zeroes of polynomial equation 4s 2 –4s+1 are (1/2, 1/2)

Sum of zeroes = (½)+(1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s 2 )

Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s 2 )

(iii) 6x 2 –3–7x

⇒6x 2 –7x–3 = 6x 2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)

Therefore, zeroes of polynomial equation 6x 2 –3–7x are (-1/3, 3/2)

Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x 2 )

Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x 2 )

(iv) 4u 2 +8u

Therefore, zeroes of polynomial equation 4u 2 + 8u are (0, -2).

Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u 2 )

Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u 2 )

(v) t 2 –15

⇒ t 2 = 15 or t = ±√15

Therefore, zeroes of polynomial equation t 2 –15 are (√15, -√15)

Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t 2 )

Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t 2 )

(vi) 3x 2 –x–4

⇒ 3x 2 –4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)

Therefore, zeroes of polynomial equation3x 2 – x – 4 are (4/3, -1)

Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x 2 )

Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x 2 )

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.

(i) 1/4 , -1

From the formulas of sum and product of zeroes, we know,

Sum of zeroes = α+β

Product of zeroes = α β

Sum of zeroes = α+β = 1/4

Product of zeroes = α β = -1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x 2 –(α+β)x +αβ = 0

x 2 –(1/4)x +(-1) = 0

4x 2 –x-4 = 0

Thus , 4x 2 –x–4 is the quadratic polynomial.

(ii) √2, 1/3

Sum of zeroes = α + β =√2

Product of zeroes = α β = 1/3

x 2  –(√2)x + (1/3) = 0

3x 2 -3√2x+1 = 0

Thus, 3x 2 -3√2x+1 is the quadratic polynomial.

(iii) 0, √5

Sum of zeroes = α+β = 0

Product of zeroes = α β = √5

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly

x 2 –(0)x +√5= 0

Thus, x 2 +√5 is the quadratic polynomial.

Sum of zeroes = α+β = 1

Product of zeroes = α β = 1

x 2 –x+1 = 0

Thus, x 2 –x+1 is the quadratic polynomial.

(v) -1/4, 1/4

Sum of zeroes = α+β = -1/4

Product of zeroes = α β = 1/4

x 2 –(-1/4)x +(1/4) = 0

4x 2 +x+1 = 0

Thus, 4x 2 +x+1 is the quadratic polynomial.

Sum of zeroes = α+β =4

Product of zeroes = αβ = 1

x 2 –(α+β)x+αβ = 0

x 2 –4x+1 = 0

Thus, x 2 –4x+1 is the quadratic polynomial.

Exercise 2.3 Page: 36

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x 3 -3x 2 +5x–3 , g(x) = x 2 –2

Dividend = p(x) = x 3 -3x 2 +5x–3

Divisor = g(x) = x 2 – 2

Therefore, upon division we get,

Quotient = x–3

Remainder = 7x–9

(ii) p(x) = x 4 -3x 2 +4x+5 , g(x) = x 2 +1-x

Dividend = p(x) = x 4 – 3x 2 + 4x +5

Divisor = g(x) = x 2 +1-x

Quotient = x 2 + x–3

Remainder = 8

(iii) p(x) =x 4 –5x+6, g(x) = 2–x 2

Dividend = p(x) =x 4 – 5x + 6 = x 4 +0x 2 –5x+6

Divisor = g(x) = 2–x 2 = –x 2 +2

Quotient = -x 2 -2

Remainder = -5x + 10

2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) t 2 -3, 2t 4 +3t 3 -2t 2 -9t-12

First polynomial = t 2 -3

Second polynomial = 2t 4 +3t 3 -2t 2 -9t-12

As we can see, the remainder is left as 0. Therefore, we say that, t 2 -3 is a factor of 2t 4 +3t 3 -2t 2 -9t-12.

(ii)x 2 +3x+1 , 3x 4 +5x 3 -7x 2 +2x+2

First polynomial = x 2 +3x+1

Second polynomial = 3x 4 +5x 3 -7x 2 +2x+2

As we can see, the remainder is left as 0. Therefore, we say that, x 2 + 3x + 1 is a factor of 3x 4 +5x 3 -7x 2 +2x+2.

(iii) x 3 -3x+1, x 5 -4x 3 +x 2 +3x+1

First polynomial = x 3 -3x+1

Second polynomial = x 5 -4x 3 +x 2 +3x+1

As we can see, the remainder is not equal to 0. Therefore, we say that, x 3 -3x+1 is not a factor of x 5 -4x 3 +x 2 +3x+1 .

3. Obtain all other zeroes of 3x 4 +6x 3 -2x 2 -10x-5, if two of its zeroes are √(5/3) and – √(5/3).

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

√(5/3) and – √(5/3) are zeroes of polynomial f(x).

∴ (x – √(5/3) ) (x+ √(5/3) = x 2 -(5/3) = 0

(3x 2 −5)=0, is a factor of given polynomial f(x).

Now, when we will divide f(x) by (3x 2 −5) the quotient obtained will also be a factor of f(x) and the remainder will be 0.

Therefore, 3x 4  +6x 3  −2x 2  −10x–5 = (3x 2  –5) (x 2 +2x+1)

Now, on further factorizing (x 2 +2x+1) we get,

x 2 +2x+1  = x 2 +x+x+1 = 0

x(x+1)+1(x+1) = 0

(x+1)(x+1) = 0

So, its zeroes are given by:  x= −1  and  x = −1.

Therefore, all four zeroes of given polynomial equation are:

√(5/3),- √(5/3) , −1 and −1.

Hence, is the answer.

4. On dividing x 3 -3x 2 +x+2 by a polynomial g(x), the quotient and remainder were x–2 and –2x+4, respectively. Find g(x).

Dividend, p(x) = x 3 -3x 2 +x+2

Quotient = x-2

Remainder = –2x+4

We have to find the value of Divisor, g(x) =?

As we know,

Dividend = Divisor × Quotient + Remainder

∴ x 3 -3x 2 +x+2 = g(x)×(x-2) + (-2x+4)

x 3 -3x 2 +x+2-(-2x+4) = g(x)×(x-2)

Therefore, g(x) × (x-2) = x 3 -3x 2 +3x-2

Now, for finding g(x) we will divide x 3 -3x 2 +3x-2 with (x-2)

Therefore, g(x) = (x 2 –x+1)

5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

(iii) deg r(x) = 0

According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;

∴ p(x) = g(x)×q(x)+r(x)

Where r(x) = 0 or degree of r(x)< degree of g(x).

Now let us proof the three given cases as per division algorithm by taking examples for each.

Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.

Let us take an example, p(x) = 3x 2 +3x+3 is a polynomial to be divided by g(x) = 3.

So, (3x 2 +3x+3)/3 = x 2 +x+1 = q(x)

Thus, you can see, the degree of quotient q(x) = 2, which also equal to the degree of dividend p(x).

Hence, division algorithm is satisfied here.

Let us take an example, p(x) = x 2  + 3 is a polynomial to be divided by g(x) = x – 1.

So, x 2  + 3 = (x – 1)×(x) + (x + 3)

Hence, quotient q(x) = x

Also, remainder r(x) = x + 3

Thus, you can see, the degree of quotient q(x) = 1, which is also equal to the degree of remainder r(x).

The degree of remainder is 0 only when the remainder left after division algorithm is constant.

Let us take an example, p(x) = x 2  + 1 is a polynomial to be divided by g(x) = x.

So, x 2  + 1 = (x)×(x) + 1

And, remainder r(x) = 1

Clearly, the degree of remainder here is 0.

Exercise 2.4 Page: 36

1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2x 3 +x 2 -5x+2; -1/2, 1, -2

Given, p(x) = 2x 3 +x 2 -5x+2

And zeroes for p(x) are = 1/2, 1, -2

∴ p(1/2) = 2(1/2) 3 +(1/2) 2 -5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0

p(1) = 2(1) 3 +(1) 2 -5(1)+2 = 0

p(-2) = 2(-2) 3 +(-2) 2 -5(-2)+2 = 0

Hence, proved 1/2, 1, -2 are the zeroes of 2x 3 +x 2 -5x+2.

Now, comparing the given polynomial with general expression, we get;

∴ ax 3 +bx 2 +cx+d = 2x 3 +x 2 -5x+2

a=2, b=1, c= -5 and d = 2

As we know, if α, β, γ are the zeroes of the cubic polynomial ax 3 +bx 2 +cx+d , then;

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α+β+γ = ½+1+(-2) = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) = -5/2 = c/a

α β γ = ½×1×(-2) = -2/2 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

(ii) x 3 -4x 2 +5x-2 ; 2, 1, 1

Given, p(x) = x 3 -4x 2 +5x-2

And zeroes for p(x) are 2,1,1.

∴ p(2)= 2 3 -4(2) 2 +5(2)-2 = 0

p(1) = 1 3 -(4×1 2 )+(5×1)-2 = 0

Hence proved, 2, 1, 1 are the zeroes of x 3 -4x 2 +5x-2

∴ ax 3 +bx 2 +cx+d = x 3 -4x 2 +5x-2

a = 1, b = -4, c = 5 and d = -2

α + β + γ = –b/a

αβ + βγ + γα = c/a

α β γ = – d/a.

α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a

αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a

αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Let us consider the cubic polynomial is ax 3 +bx 2 +cx+d and the values of the zeroes of the polynomials be α, β, γ.

As per the given question,

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

Thus, from above three expressions we get the values of coefficient of polynomial.

a = 1, b = -2, c = -7, d = 14

Hence, the cubic polynomial is x 3 -2x 2 -7x+14

3. If the zeroes of the polynomial x 3 -3x 2 +x+1 are a – b, a, a + b, find a and b.

We are given with the polynomial here,

p(x) = x 3 -3x 2 +x+1

And zeroes are given as a – b, a, a + b

∴px 3 +qx 2 +rx+s = x 3 -3x 2 +x+1

p = 1, q = -3, r = 1 and s = 1

Sum of zeroes = a – b + a + a + b

Putting the values q and p.

-(-3)/1 = 3a

Thus, the zeroes are 1-b, 1, 1+b.

Now, product of zeroes = 1(1-b)(1+b)

-s/p = 1-b 2

-1/1 = 1-b 2

b 2 = 1+1 = 2

Hence,1-√2, 1 ,1+√2 are the zeroes of x 3 -3x 2 +x+1.

4. If two zeroes of the polynomial x 4 -6x 3 -26x 2 +138x-35 are 2 ± √ 3, find other zeroes.

Let f(x) = x 4 -6x 3 -26x 2 +138x-35

Since 2 +√ 3 and 2-√ 3 are zeroes of given polynomial f(x).

∴ [x−(2+√ 3 )] [x−(2-√ 3) ] = 0

(x−2−√ 3 )(x−2+√ 3 ) = 0

On multiplying the above equation we get,

x 2 -4x+1, this is a factor of a given polynomial f(x).

Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

So, x 4 -6x 3 -26x 2 +138x-35 = (x 2 -4x+1)(x 2  –2x−35)

Now, on further factorizing (x 2 –2x−35) we get,

x 2 –(7−5)x −35  = x 2 – 7x+5x+35 = 0

x(x −7)+5(x−7) = 0

(x+5)(x−7) = 0

So, its zeroes are given by:

x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are: 2+√ 3 , 2-√ 3 , −5 and 7.

Q.5: If the polynomial x 4 – 6x 3 + 16x 2 – 25x + 10 is divided by another polynomial x 2 – 2x + k, the remainder comes out to be x + a, find k and a.

Let’s divide x 4 – 6x 3 + 16x 2 – 25x + 10 by x 2 – 2x + k.

Given that the remainder of the polynomial division is x + a.

(4k – 25 + 16 – 2k)x + [10 – k(8 – k)] = x + a

(2k – 9)x + (10 – 8k + k 2 ) = x + a

Comparing the coefficients of the above equation, we get;

2k = 9 + 1 = 10

k = 10/2 = 5

10 – 8k + k 2 = a

10 – 8(5) + (5) 2 = a [since k = 5]

10 – 40 + 25 = a

Therefore, k = 5 and a = -5.

NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials

As this is one of the important topics in Maths, it comes under the unit – Algebra which has a weightage of 20 marks in the Class 10 Maths CBSE exams. The average number of questions asked from this chapter is usually 1. This chapter talks about the following,

• Introduction to Polynomials
• Geometrical Meaning of the Zeros of Polynomial
• Relationship between Zeros and Coefficients of a Polynomial
• Division Algorithm for Polynomials

Polynomials are introduced in Class 9, where we discussed polynomials in one variable and their degrees in the previous class. This is discussed in further detail in Class 10. The NCERT Solutions for Class 10 Maths for this chapter discusses the answers to various types of questions related to polynomials and their applications. We study the division algorithm for polynomials of integers, and also whether the zeroes of quadratic polynomials are related to their coefficients.

The chapter starts with the introduction of polynomials in section 2.1, followed by two very important topics in sections 2.2 and 2.3

• Geometrical Meaning of the zeroes of a Polynomial – It includes 1 question having 6 different cases.
• Relationship between Zeroes and Coefficients of a Polynomial – Explore the relationship between zeroes and coefficients of a quadratic polynomial through solutions to 2 problems in Exercise 2.2, having 6 parts in each question.

Next, it discusses the following topics, which were introduced in Class 9.

• Division Algorithm for Polynomials – In this, the solutions for 5 problems in Exercise 2.3 is given, having three long questions.

Key Features of NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials

• It covers the CBSE syllabus for 2023-24 of Class 10 Maths.
• After studying these NCERT Solutions prepared by our subject experts, you will be confident of scoring well in the exams.
• It follows NCERT guidelines which help in preparing the students accordingly.
• It contains all the important questions from the examination point of view.

For a strong grip over the concepts, students can also make use of the other reference materials which are present at BYJU’S.

• RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials

Disclaimer – 

Dropped Topics –  2.4 Division algorithm for polynomials

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