---
title: "NCERT Solutions for Class 10 Maths Chapter 2 Polynomials"
url: https://www.swavid.com/maths/class/10/chapter/polynomials/ncert-solutions
dateModified: 2026-10-07T15:48:01+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

This chapter's questions cover finding the number of zeroes of polynomials from their graphs, finding the zeroes of quadratic polynomials, and verifying the relationship between zeroes and coefficients.

Free PDF (5 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-2-polynomials-68994c428f.pdf

## EXERCISE 2.1

### Question 1

*6 marks · Short answer*

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.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi)

**Part (i)**

1. The graph of $y = p(x)$ intersects the $x$-axis at exactly one point.
2. Therefore, the number of zeroes is 1.

Answer (i): 1

**Part (ii)**

1. The graph of $y = p(x)$ intersects the $x$-axis at exactly one point.
2. Therefore, the number of zeroes is 1.

Answer (ii): 1

**Part (iii)**

1. The graph of $y = p(x)$ intersects the $x$-axis at three distinct points.
2. Therefore, the number of zeroes is 3.

Answer (iii): 3

**Part (iv)**

1. The graph of $y = p(x)$ intersects the $x$-axis at two distinct points.
2. Therefore, the number of zeroes is 2.

Answer (iv): 2

**Part (v)**

1. The graph of $y = p(x)$ intersects the $x$-axis at four points.
2. Therefore, the number of zeroes is 4.

Answer (v): 4

**Part (vi)**

1. The graph of $y = p(x)$ intersects or touches the $x$-axis at three points.
2. Therefore, the number of zeroes is 3.

Answer (vi): 3

**Answer:** The number of zeroes in cases (i) to (vi) are 1, 1, 3, 2, 4, and 3 respectively.

> Common mistake: Counting the number of times the graph intersects the y-axis or counting local maxima and minima instead of intersection points with the x-axis.

## EXERCISE 2.2

### Question 1

*3 marks · Short answer*

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^2 - 2x - 8$
(ii) $4s^2 - 4s + 1$
(iii) $6x^2 - 3 - 7x$
(iv) $4u^2 + 8u$
(v) $t^2 - 15$
(vi) $3x^2 - x - 4$

**Part (i)**

1. $x^2 - 2x - 8 = (x - 4)(x + 2)$
2. The zeroes are $4$ and $-2$
3. Sum of zeroes = $4 + (-2) = 2 = -\frac{-2}{1}$
4. Product of zeroes = $4 \times (-2) = -8 = \frac{-8}{1}$

Answer (i): Zeroes: $4, -2$

**Part (ii)**

1. $4s^2 - 4s + 1 = (2s - 1)^2$
2. The zeroes are $\frac{1}{2}$ and $\frac{1}{2}$
3. Sum = $\frac{1}{2} + \frac{1}{2} = 1 = -\frac{-4}{4}$
4. Product = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{1}{4}$

Answer (ii): Zeroes: $\frac{1}{2}, \frac{1}{2}$

**Part (iii)**

1. $6x^2 - 7x - 3 = (3x + 1)(2x - 3)$
2. The zeroes are $-\frac{1}{3}$ and $\frac{3}{2}$
3. Sum = $-\frac{1}{3} + \frac{3}{2} = \frac{7}{6} = -\frac{-7}{6}$
4. Product = $(-\frac{1}{3}) \times \frac{3}{2} = -\frac{1}{2} = \frac{-3}{6}$

Answer (iii): Zeroes: $-\frac{1}{3}, \frac{3}{2}$

**Part (iv)**

1. $4u^2 + 8u = 4u(u + 2)$
2. The zeroes are $0$ and $-2$
3. Sum = $0 + (-2) = -2 = -\frac{8}{4}$
4. Product = $0 \times (-2) = 0 = \frac{0}{4}$

Answer (iv): Zeroes: $0, -2$

**Part (v)**

1. $t^2 - 15 = (t - \sqrt{15})(t + \sqrt{15})$
2. The zeroes are $\sqrt{15}$ and $-\sqrt{15}$
3. Sum = $\sqrt{15} - \sqrt{15} = 0 = -\frac{0}{1}$
4. Product = $\sqrt{15} \times (-\sqrt{15}) = -15 = \frac{-15}{1}$

Answer (v): Zeroes: $\sqrt{15}, -\sqrt{15}$

**Part (vi)**

1. $3x^2 - x - 4 = (3x - 4)(x + 1)$
2. The zeroes are $\frac{4}{3}$ and $-1$
3. Sum = $\frac{4}{3} - 1 = \frac{1}{3} = -\frac{-1}{3}$
4. Product = $\frac{4}{3} \times (-1) = -\frac{4}{3} = \frac{-4}{3}$

Answer (vi): Zeroes: $\frac{4}{3}, -1$

**Answer:** Zeroes and relationship verified for all six quadratic polynomials.

> Common mistake: Incorrect signs while splitting the middle term or writing coefficients.

### Question 2

*3 marks · Short answer*

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) $\frac{1}{4}, -1$
(ii) $\sqrt{2}, \frac{1}{3}$
(iii) $0, \sqrt{5}$
(iv) $1, 1$
(v) $-\frac{1}{4}, \frac{1}{4}$
(vi) $4, 1$

**Part (i)**

1. Let the quadratic polynomial be $ax^2 + bx + c$ and its zeroes be $\alpha$ and $\beta$.
2. Given sum of zeroes $\alpha + \beta = \frac{1}{4}$ and product of zeroes $\alpha\beta = -1$.
3. A quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha\beta] = k\left[x^2 - \frac{1}{4}x - 1\right]$.
4. For $k = 4$, the polynomial is $4x^2 - x - 4$.

Answer (i): $x^2 - \frac{1}{4}x - 1$ (or $4x^2 - x - 4$)

**Part (ii)**

1. Given sum of zeroes $\alpha + \beta = \sqrt{2}$ and product of zeroes $\alpha\beta = \frac{1}{3}$.
2. A quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha\beta] = k\left[x^2 - \sqrt{2}x + \frac{1}{3}\right]$.
3. For $k = 3$, the polynomial is $3x^2 - 3\sqrt{2}x + 1$.

Answer (ii): $x^2 - \sqrt{2}x + \frac{1}{3}$ (or $3x^2 - 3\sqrt{2}x + 1$)

**Part (iii)**

1. Given sum of zeroes $\alpha + \beta = 0$ and product of zeroes $\alpha\beta = \sqrt{5}$.
2. A quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha\beta] = k[x^2 - 0x + \sqrt{5}]$.
3. For $k = 1$, the polynomial is $x^2 + \sqrt{5}$.

Answer (iii): $x^2 + \sqrt{5}$

**Part (iv)**

1. Given sum of zeroes $\alpha + \beta = 1$ and product of zeroes $\alpha\beta = 1$.
2. A quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha\beta] = k[x^2 - x + 1]$.
3. For $k = 1$, the polynomial is $x^2 - x + 1$.

Answer (iv): $x^2 - x + 1$

**Part (v)**

1. Given sum of zeroes $\alpha + \beta = -\frac{1}{4}$ and product of zeroes $\alpha\beta = \frac{1}{4}$.
2. A quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha\beta] = k\left[x^2 - \left(-\frac{1}{4}\right)x + \frac{1}{4}\right]$.
3. For $k = 4$, the polynomial is $4x^2 + x + 1$.

Answer (v): $x^2 + \frac{1}{4}x + \frac{1}{4}$ (or $4x^2 + x + 1$)

**Part (vi)**

1. Given sum of zeroes $\alpha + \beta = 4$ and product of zeroes $\alpha\beta = 1$.
2. A quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha\beta] = k[x^2 - 4x + 1]$.
3. For $k = 1$, the polynomial is $x^2 - 4x + 1$.

Answer (vi): $x^2 - 4x + 1$

**Answer:** Quadratic polynomials found for all six cases.

> Common mistake: Swapping the signs of the sum of zeroes while substituting into the general formula $x^2 - (\alpha + \beta)x + \alpha\beta$.

## Frequently asked questions

### How many exercises and questions are there in NCERT Solutions for Class 10 Maths Chapter 2 Polynomials for the 2026-27 session?

This chapter contains a total of two exercises with 3 questions in all. Exercise 2.1 has 1 short answer question, while Exercise 2.2 has 2 short answer questions. You can find step-by-step solutions for all of them in SwaVid's free PDF available on this page.

### Which mathematical topics and concepts do the questions cover in this chapter?

The questions cover the geometrical meaning of the zeroes of a polynomial, finding zeroes of quadratic polynomials, and verifying the relationship between zeroes and coefficients. They also include forming a quadratic polynomial when the sum and product of its zeroes are given. SwaVid's free PDF on this page explains all these concepts clearly.

### Which is the hardest question type in this chapter and how should students approach it?

The most challenging questions involve verifying the relationship between zeroes and coefficients or finding polynomials using given roots. To approach these, first find the zeroes using factorization like $ax^2 + bx + c = 0$ and then apply standard formulas. SwaVid provides detailed methods for these problems in the solutions on this page.

### How can students write answers in exams to score full marks for Class 10 Polynomials?

To score full marks, you should write every step clearly, state the formulas used, and show the proper substitution of values. Drawing clear graphs for geometrical meaning and double-checking coefficient verifications helps prevent silly mistakes. You can refer to SwaVid's free PDF on this page to learn the correct presentation style.

### Is the free PDF for NCERT Solutions of Class 10 Maths Chapter 2 available on this page?

Yes, the complete free PDF and step-by-step solutions for this chapter are available on this page only. It is fully updated according to the NCERT textbook for the 2026-27 session. You can easily download it to practice all the textbook exercises offline.

## Related pages

- [Exercise 2.1 solutions](https://www.swavid.com/maths/class/10/chapter/polynomials/ncert-solutions/exercise-2-1)
- [Exercise 2.2 solutions](https://www.swavid.com/maths/class/10/chapter/polynomials/ncert-solutions/exercise-2-2)
- [Polynomials: CBSE previous year questions](https://www.swavid.com/cbse/class-10/maths/pyq/polynomials)
- [Class 10 Maths chapters](https://www.swavid.com/maths/class/10)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student