# NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2

Question 1.
Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0
(ii) x = -1
(iii) x = 2
Solution:
(i) Let P(x) = 5x – 4x2 + 3
Therefore, P(0) = 5(0) – 4(0)2 + 3
= 0 – 0 + 3
= 3
So, the value of P(x) at x = 0 is 3.

(ii) Let P(x) = 5x – 4x2 + 3
Therefore, P(-1) = 5(-1) – 4(-1)2 + 3
= -5 – 4 + 3
= -6
So, the value of P(x) at x = -1 is -6.

(iii) Let P(x) = 5x – 4x2 + 3
Therefore, P(2) = 5(2) – 4(2)2 + 3
= 10 – 16 + 3
= -3
So, the value of P(x) at x = 2 is -3.

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y2 – y + 1
(ii) p(t) = 2 + t + 2t2 – t3
(iii) p(x) = x3
(iv) p(x) = (x – 1) (x + 1)
Solution:
(i) We have given
p(y) = y2 – y + 1
Therefore, the value of polynomial p(y) at y = 0 is
p(0) = 02 – 0 + 1 = 1
Again, the value of polynomial p(y) at y = 1 is
p(1) = 12 – 1 + 1 = 1
Again, the value of polynomial p(y) at y = 2 is
p(2) = 22 – 2 + 1 = 3

(ii) We have given,
p(t) = 2 + t + 2t2 – t3
Therefore, the value of polynomial p(t) at t = 0 is
p(0) = 2 + 0 + 2(0)2 – (0)3 = 2
Again, the value of polynomial p(t) at t = 1 is
p(1) = 2 + 1 + 2(1)2 – (1)3 = 4
Again, the value of polynomial p(t) at t = 2 is
p(2) = 2 + 2 + 2(2)2 – (2)3 = 4

(iii) We have given,
P(x) = x3,
Therefore, the value of polynomial p(x) at x = 0 is
p(0) = (0)3 = 0
Again the value of polynomial p(x) at x = 1 is
p(1) = (1)3 = 1
Again, the value of polynomial p(x) at x = 2 is
P(2) = (2)3 = 8

(iv) We have given
p(x) = (x – 1)(x + 1)
Therefore, the value of polynomial p(x) at x = 0 is
P(0) = (0 – 1)(0 + 1)
= (-1) × (+1)
= -1
Again, the value of polynomial p(x) at x = 1 is
p(1) = (1 – 1) × (1 + 1)
= 0 × 2
= 0
Again, the value of polynomial p(x), at x = 2 is
p(2) = (2 – 1)(2 + 1)
= 1 × 3
= 3

Question 3.
Verify whether the following are zeros of the polynomial indicated against them:
(i) p(x) = 3x + 1, x = $$-\frac{1}{3}$$
(ii) p(x) = 5x – p, x = $$\frac{4}{5}$$
(iii) p(x) = x2 – 1, x = 1, -1
(iv) p(x) = (x + 1) (x – 2), x = -1, 2
(v) p(x) = x2, x = 0
(vi) p(x) = 1x + m, x = $$-\frac{m}{1}$$
(vii) p(x) = 3x2 – 1, x = $$-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$$
(viii) p(x) = 2x + 1, x = $$\frac{1}{2}$$
Solution:
(i) We have given that, p(x) = 3x + 1.
Therefore, the value of polynomial p(x) at x = $$-\frac{1}{3}$$ is
$$P\left(-\frac{1}{3}\right)=3 \times \frac{(-1)}{3}+1$$ = 0
yes x = $$-\frac{1}{3}$$ is the zero of polynomial p(x).

(ii) We have given that p(x) = 5x – p
Therefore, the value of polynomial p(x) at x = $$\frac{4}{5}$$ is
$$P\left(\frac{4}{5}\right)=5 \times \frac{4}{5}-\frac{22}{7}$$ (∵ π = $$\frac {22}{7}$$)
= 4 – $$\frac{22}{7}$$
$$P\left(\frac{4}{5}\right)=\frac{6}{7}$$
No, x = $$\frac{4}{5}$$ is not the zero of p(x) = 5x – π.

(iii) We have given that p(x) = x2 – 1
Therefore, the value of polynomial p(x) at x = 1 is
p(1) = 12 – 1 = 0
Again, the value of polynomial p(x) at x = -1 is
p(-1) = (-1)2 – 1 = 0
yes x = 1, -1 are the zero of polynomial p(x) = x2 – 1

(iv) We have given that
p(x) = (x + 1)(x – 2)
Therefore, the value of polynomial p(x) at x = -1 is
p(-1) = (-1 + 1)(-1 – 2)
= 0 × (-3)
= 0
Again, the value of polynomial p(x) at x = 2 is
P(2) = (2 + 1)(2 – 2)
= 3 × 0
= 0
yes x = -1, 2 are the zero of polynomial p(x) = (x + 1)(x – 2)

(v) We have given that p(x) = x2
Therefore, the value of polynomial p(x) at x = 0 is
p(0) = 02 = 0
yes x = 0 is the zero of polynomial p(x) = x2.

(vi) We have given that, p(x) = lx + m
Therefore, the value of polynomial p(x) at x = $$-\frac{m}{\ell}$$ is
$$P\left(-\frac{m}{\ell}\right)=\ell\left(-\frac{m}{\ell}\right)+m$$ = 0
yes x = $$-\frac{m}{\ell}$$ is the zero of polynomial p(x) = lx + m.

(vii) We have given that, p(x) = 3x2 – 1
Therefore, the value of polynomial p(x) at x = $$-\frac{1}{\sqrt{3}}$$ is
$$P\left(-\frac{1}{\sqrt{3}}\right)=3 \times\left(-\frac{1}{\sqrt{3}}\right)^{2}-1$$
= $$3 \times \frac{1}{3}-1$$
= 0
Again, the value of polynomial p(x) at x = $$\frac{2}{\sqrt{3}}$$ is
$$P\left(\frac{2}{\sqrt{3}}\right)=3 \times\left(\frac{2}{\sqrt{3}}\right)^{2}-1$$
= $$3 \times \frac{4}{3}-1$$
= 4 – 1
= 3
Therefore, x = $$-\frac{1}{\sqrt{3}}$$ is the zero of polynomial p(x) and x = $$\frac{2}{\sqrt{3}}$$ is not the zero of polynomial p(x).

(viii) We have given that p(x) = 2x + 1
Therefore, the value of polynomial p(x) at x = $$\frac{1}{2}$$ is
$$\mathrm{P}\left(\frac{1}{2}\right)=2 \times \frac{1}{2}+1$$ = 2
Therefore, x = $$\frac{1}{2}$$ is not the zero of polynomial p(x).

Question 4.
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
(ii) p(x) = x – 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2
(v) p(x) = 3x
(vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0; c, d are real numbers.
Solution:
(i) We have given that
p(x) = x + 5 ……..(i)
To find the zero of polynomial p(x), we can take
P(x) = 0 ………(ii)
from equation (i) and (ii)
x + 5 = 0
∴ x = -5
Therefore, x = -5 is the zero of polynomial p(x) = x + 5.

(ii) We have given that,
p(x) = x – 5 ……..(i)
To find the zero of polynomial p(x) we can take
p(x) = 0 ……..(ii)
from equation (i) and (ii)
x – 5 = 0
∴ x = 5
Therefore x = 5 is the zero of polynomial p(x) = x – 5.

(iii) We have given that
p(x) = 2x + 5 ………(i)
To find the zero of polynomial p(x), we can take
p(x) = 0 …….(ii)
from equation (i) and (ii)
2x + 5 = 0
or x = $$-\frac{5}{2}$$
Therefore x = $$-\frac{5}{2}$$ is the zero of polynomial p(x) = 2x + 5.

(iv) We have given that
p(x) = 3x – 2 …….(i)
To find the zero of polynomial p(x) we can take
p(x) = 0 ……..(ii)
3x – 2 = 0
x = $$\frac{2}{3}$$
Therefore, x = $$\frac{2}{3}$$ is the zero of polynomial p(x) = 3x – 2.

(v) We have given that
p(x) = 3x ……..(i)
To find the zero of polynomial p(x) we can take
P(x) = 0 ……..(ii)
from equation (i) and (ii)
3x = 0
x = 0
Therefore, x = 0 is the zero of polynomial p(x) = 3x.

(vi) We have given that,
p(x) = ax, a ≠ 0 ……(i)
To find the zero of polynomial p(x) we can take
p(x) = 0 ………(ii)
from equation (i) and (ii)
ax = 0
x = $$\frac{0}{a}$$ = 0
Therefore, x = 0 is the zero of polynomial p(x) = ax, a ≠ 0.

(vii) We have given
p(x) = cx + d …….(i)
c ≠ 0, c, d are real numbers.
To find the zero of polynomial p(x) we can take
p(x) = 0 ………(ii)
from equation (i) and (ii)
cx + d = 0
x = $$-\frac{d}{c}$$
where c ≠ 0 and c, d are real numbers.
Therefore, x = $$\frac{d}{c}$$, where c ≠ 0 and c, d are real numbers is the solution of polynomial p(x) = cx + d.

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