Exercise 2.2 NCERT Class 9 Polynomials Best Solutions

Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 Exercise 2.1 Exercise 2.2 Exercise 2.3 Exercise 2.4

Exercise 2.2 NCERT Class 9 contains a total of four questions. The questions are based on the following topics.

Zeroes of a Polynomial
Roots of a Polynomial
Value of a Polynomial at a point
Verifying a zero of a Polynomial

Exercise 2.2 NCERT Class 9 Polynomials Solutions

Q.1 Q.2 Q.3 Q.4 Q.5

The first question in Exercise 2.2 NCERT Class 9 is about finding the value of the polynomial at a point.

1. Find the value of the polynomial $5 x - 4 x^{2} + 3$ at
(i) $x = 0$
(ii) $x = - 1$
(iii) $x = 2$

Answer

The given polynomial is $5 x - 4 x^{2} + 3$ .

(i) putting $x = 0$ in the polynomial, we get.
$5 x - 4 x^{2} + 3$
= $5 (0) - 4 {(0)}^{2} + 3$
= $0 - 0 + 3$
= 3

(ii) putting $x = - 1$ in the polynomial, we get.
$5 x - 4 x^{2} + 3$
$= 5 (- 1) - 4 {(- 1)}^{2} + 3$
$= - 5 - 4 \times 1 + 3$
$= - 6$

(iii) putting $x = 2$ in the polynomial, we get.
$5 x - 4 x^{2} + 3$
$= 5 (2) - 4 {(2)}^{2} + 3$
$= 10 - 4 \times 4 + 3$
$= 10 - 16 + 3$
$= - 3$

The second question of Exercise 2.2 NCERT Class 9 asks us to find value of a polynomial at a point.

2. Find $p (0)$ , $p (1)$ and $p (2)$ for each of the following polynomials:

(i) $p (y) = y^{2} - y + 1$

Answer

The given polynomial is
$p (y) = y^{2} - y + 1$

Putting $y = 0$ , we get.
$p (0) = 0^{2} - 0 + 1$
$= 0 - 0 + 1$
$= 1$ .

Putting $y = 1$ , we get.
$p (1) = 1^{2} - 1 + 1$
$= 1 - 1 + 1$
$= 1$

Putting $y = 2$ , we get.
$p (2) = 2^{2} - 2 + 1$
$= 4 - 2 + 1$
$= 3$

(ii) $p (t) = 2 + t + 2 t^{2} - t^{3}$

Answer

The given polynomial is
$p (t) = 2 + t + 2 t^{2} - t^{3}$

Putting $t = 0$ , we get.
$p (0) = 2 + 0 + 2 {(0)}^{2} - 0^{3}$
$= 2 + 0 + 0 - 0$
$= 2$

Putting $t = 1$ , we get.
$p (1) = 2 + 1 + 2 {(1)}^{2} - 1^{3}$
$= 2 + 1 + 2 - 1$
$= 4$

Putting $t = 2$ , we get.
$p (2) = 2 + 2 + 2 {(2)}^{2} - 2^{3}$
$= 2 + 2 + 8 - 8$
$= 4$

(iii) $p (x) = x^{3}$

Answer

The given polynomial is
$p (x) = x^{3}$

Putting $x = 0$ , we get.
$p (0) = 0^{3}$
$= 0$ .

Putting $x = 1$ , we get.
$p (1) = 1^{3}$
$= 1$ .

Putting $x = 2$ , we get.
$p (2) = 2^{3}$
$= 8$ .

(iv) $p (x) = (x - 1) (x + 1)$

Answer

The given polynomial is
$p (x) = (x - 1) (x + 1)$

Putting $x = 0$ , we get.
$p (0) = (0 - 1) (0 + 1)$
$= (- 1) \times 1$
$= - 1$

Putting $x = 1$ , we get.
$p (1) = (1 - 1) (1 + 1)$
$= 0 \times 2$
$= 0$

Putting $x = 2$ , we get.
$p (2) = (2 - 1) (1 + 1)$
$= 1 \times 2$
$= 2$

In Exercise 2.2 NCERT Class 9, the third question asks us to verify whether the given zeroes are zeroes of the polynomial, indicated against them.

3. Verify whether the following are zeroes of the polynomial, indicated against them.

(i) $p (x) = 3 x + 1, x = - \frac{1}{3}$

Answer

The given polynomial is $p (x) = 3 x + 1$ .
Putting $x = - \frac{1}{3}$ , we get.

$p (- \frac{1}{3}) = 3 (- \frac{1}{3}) + 1$ = $- 1 + 1$ = 0

$∵ p (\frac{- 1}{3}) = 0$ , $x = - \frac{1}{3}$ is a zero of the given polynomial.

(ii) $p (x) = 5 x - π, x = \frac{4}{5}$

Answer

We have
$p (x) = 5 x - π$

Putting $x = \frac{4}{5}$ , we get.

$p (\frac{4}{5}) = 5 \times \frac{4}{5} - π = 4 - π$

$∵ p (\frac{4}{5}) \neq 0$ , $x = \frac{4}{5}$ is not a zero of the given polynomial.

(iii) $p (x) = x^{2} - 1, x = 1, - 1$

Answer

We have
$p (x) = x^{2} - 1$

Putting $x = 1$ , we get.
$p (1) = 1^{2} - 1$ $= 1 - 1$ $= 0$
$∵ p (1) = 0$ , 1 is a zero of $p (x)$ .

Putting $x = −1$ , we get.
$p (- 1) = {(- 1)}^{2} - 1$ $= 1 - 1$ $= 0$
$∵ p (- 1) = 0$ , −1 is a zero of $p (x)$ .

(iv) $p (x) = (x + 1) (x - 2), x = - 1, 2$

Answer

We have
$p (x) = (x + 1) (x - 2)$
Putting $x = - 1$ , we get.
$p (- 1) = (- 1 + 1) (- 1 - 2)$ $= 0 \times (- 3) = 0$
$∵ p (- 1) = 0,$ −1 is a zero of $p (x)$ .

Putting $x = 2$ , we get.
$p (2) = (2 + 1) (2 - 2)$ $= 3 \times 0$ $= 0$ .
$∵ p (2) = 0$ , 2 is a zero of $p (x)$ .

(v) $p (x) = x^{2}, x = 0$

Answer

We have
$p (x) = x^{2}$

Putting $x = 0$ , we get.
$p (0) = 0^{2} = 0$ .
$∵ p (0) = 0$ , 0 is a zero of $p (x)$ .

(vi) $p (x) = l x + m, x = - \frac{m}{l}$

Answer

We have
$p (x) = l x + m$

Putting $x = - \frac{m}{l}$ , we get.
$p (- \frac{m}{l}) = l \times (- \frac{m}{l}) + m$ $= - m + m$ $= 0$
$∵ p (- \frac{m}{l}) = 0$ , $- \frac{m}{l}$ is a zero of $p (x)$ .

(vii) $p (x) = 3 x^{2} - 1, x = - \frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$

Answer

We have,
$p (x) = 3 x^{2} - 1$

Putting $x = - \frac{1}{\sqrt{3}}$ , we get.

$p (- \frac{1}{\sqrt{3}}) = 3 {(- \frac{1}{\sqrt{3}})}^{2} - 1$ $= 3 \times \frac{1}{3} - 1$ $= 1 - 1$ $= 0$

$∵ p (- \frac{1}{\sqrt{3}}) = 0$ , $- \frac{1}{\sqrt{3}}$ is a zero of $p (x)$ .

Putting $x = \frac{2}{\sqrt{3}}$ , we get.

$p (\frac{2}{\sqrt{3}}) = 3 \times {(\frac{2}{\sqrt{3}})}^{2} - 1$ $= 3 \times \frac{4}{3} - 1 = 4 - 1 = 3$

$∵ p (\frac{2}{\sqrt{3}}) \neq 0$ , $\frac{2}{\sqrt{3}}$ is not a zero of $p (x)$ .

(viii) $p (x) = 2 x + 1, x = \frac{1}{2}$

Answer

We have
$p (x) = 2 x + 1$

Putting $x = \frac{1}{2}$ , we get.

$p (\frac{1}{2}) = 2 \times \frac{1}{2} + 1 = 1 + 1 = 2$

$∵ p (\frac{1}{2}) \neq 0$ , $\frac{1}{2}$ is not a zero of $p (x)$ .

In Exercise 2.2 NCERT Class 9, the fourth question asks us to find out the zero of the given polynomial.

4. Find the zero of the polynomial in each of the following cases:

(i) $p (x) = x + 5$

Answer

Let $' a'$ be a zero of $p (x)$ .
$∴ p (a) = 0$
$\Rightarrow a + 5 = 0$
$\Rightarrow a = −5$ .
Hence, −5 is the zero of the polynomial $p (x)$ .

(ii) $p (x) = x - 5$

Answer

Let $' a'$ be a zero of $p (x)$ .
$∴ p (a) = 0$
$\Rightarrow a - 5 = 0$
$\Rightarrow a = 5$ .
Hence, 5 is the zero of the polynomial $p (x)$ .

(iii) $p (x) = 2 x + 5$

Answer

Let $' a'$ be a zero of $p (x)$ .
$∴ p (a) = 0$
$\Rightarrow 2 a + 5 = 0$
$\Rightarrow 2 a = −5$ .
$\Rightarrow a = - \frac{5}{2}$
Hence, $- \frac{5}{2}$ is the zero of the polynomial $p (x)$ .

(iv) $p (x) = 3 x - 2$

Answer

Let $' a'$ be a zero of $p (x)$ .
$∴ p (a) = 0$
$\Rightarrow 3 a - 2 = 0$
$\Rightarrow 3 a = 2$ .
$\Rightarrow a = \frac{2}{3}$
Hence, $\frac{2}{3}$ is the zero of the polynomial $p (x)$ .

(v) $p (x) = 3 x$

Answer

Let $' a'$ be a zero of $p (x)$ .
$∴ p (a) = 0$
$\Rightarrow 3 a = 0$
$\Rightarrow a = 0$
Hence, $0$ is the zero of the polynomial $p (x)$ .

(vi) $p (x) = a x, a \neq 0$

Answer

Let $' y'$ be a zero of $p (x)$ .
$∴ p (y) = 0$
$\Rightarrow a y = 0$
$\Rightarrow y = \frac{0}{a} = 0$
Hence, $0$ is the zero of the polynomial $p (x)$ .

(vi) $p (x) = c x + d, c \neq 0, c, d a r e r e a l n u m b e r s .$

Answer

Let $' a'$ be a zero of $p (x)$ .
$∴ p (a) = 0$
$\Rightarrow c a + d = 0$
$\Rightarrow c a = −d$
$\Rightarrow a = - \frac{d}{c}$
Hence, $- \frac{d}{c}$ is the zero of the polynomial $p (x)$ .

Important Links

CBSE Class 9 Sample Papers 2025-26

Download NCERT Book Class 9th

Buy RS Aggarwal Class 9 Book

Chapter Test ML Aggarwal Class 9 Rational and Irrational Numbers Full Solutions
Read More »

07/07/2025

Exercise 1.5 ML Aggarwal Class 9 ICSE Mathematics Chapter 1 Rational and Irrational Numbers Best Solutions
Read More »

04/07/2025

Exercise 1.4 ML Aggarwal Class 9 ICSE Mathematics Chapter 1 Rational and Irrational Numbers Best Solutions
Read More »

01/07/2025

Polynomial Worksheets Class 9 Mathematics Best for Practice
Read More »

30/06/2025

Exercise 1.3 ML Aggarwal Class 9 ICSE Mathematics Chapter 1 Rational and Irrational Numbers Best Solutions
Read More »

25/06/2025

Motion NCERT Class 9 Concise Notes and Best Question Answers
Read More »

16/06/2025

Exercise 2.2 NCERT Class 9 Polynomials Solutions

1. Find the value of the polynomial 5x−4x2+3 at (i) x=0 (ii) x=-1 (iii) x=2

Answer

2. Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y)=y2−y+1

Answer

(ii) p(t)=2+t+2t2−t3

Answer

(iii) p(x)=x3

Answer

(iv) p(x)=(x-1)(x+1)

Answer

3. Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x)=3x+1, x=−13

Answer

(ii) p(x)=5x−π, x=45

Answer

(iii) p(x)=x2−1, x=1,−1

Answer

(iv) p(x)=(x+1)(x−2), x=−1,2

Answer

(v) p(x)=x2, x=0

Answer

(vi) p(x)=lx+m, x=−ml

Answer

(vii) p(x)=3x2−1, x=−13,23

Answer

(viii) p(x)=2x+1, x=12

Answer

4. Find the zero of the polynomial in each of the following cases:

(i) p(x)=x+5

Answer

(ii) p(x)=x−5

Answer

(iii) p(x)=2x+5

Answer

(iv) p(x)=3x−2

Answer

(v) p(x)=3x

Answer

(vi) p(x)=ax, a≠0

Answer

(vi) p(x)=cx+d, c≠0,c,d are real numbers.

Answer

Important Links

1. Find the value of the polynomial $5 x - 4 x^{2} + 3$ at
(i) $x = 0$
(ii) $x = - 1$
(iii) $x = 2$

2. Find $p (0)$ , $p (1)$ and $p (2)$ for each of the following polynomials:

(i) $p (y) = y^{2} - y + 1$

(ii) $p (t) = 2 + t + 2 t^{2} - t^{3}$

(iii) $p (x) = x^{3}$

(iv) $p (x) = (x - 1) (x + 1)$

(i) $p (x) = 3 x + 1, x = - \frac{1}{3}$

(ii) $p (x) = 5 x - π, x = \frac{4}{5}$

(iii) $p (x) = x^{2} - 1, x = 1, - 1$

(iv) $p (x) = (x + 1) (x - 2), x = - 1, 2$

(v) $p (x) = x^{2}, x = 0$

(vi) $p (x) = l x + m, x = - \frac{m}{l}$

(vii) $p (x) = 3 x^{2} - 1, x = - \frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$

(viii) $p (x) = 2 x + 1, x = \frac{1}{2}$

(i) $p (x) = x + 5$

(ii) $p (x) = x - 5$

(iii) $p (x) = 2 x + 5$

(iv) $p (x) = 3 x - 2$

(v) $p (x) = 3 x$

(vi) $p (x) = a x, a \neq 0$

(vi) $p (x) = c x + d, c \neq 0, c, d a r e r e a l n u m b e r s .$