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 – 4x^{2} + 3 at

(i) x = 0

(ii) x = -1

(iii) x = 2

Solution:

(i) Let P(x) = 5x – 4x^{2} + 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 – 4x^{2} + 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 – 4x^{2} + 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) = y^{2} – y + 1

(ii) p(t) = 2 + t + 2t^{2} – t^{3}

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

(iv) p(x) = (x – 1) (x + 1)

Solution:

(i) We have given

p(y) = y^{2} – y + 1

Therefore, the value of polynomial p(y) at y = 0 is

p(0) = 0^{2} – 0 + 1 = 1

Again, the value of polynomial p(y) at y = 1 is

p(1) = 1^{2} – 1 + 1 = 1

Again, the value of polynomial p(y) at y = 2 is

p(2) = 2^{2} – 2 + 1 = 3

(ii) We have given,

p(t) = 2 + t + 2t^{2} – t^{3}

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) = x^{3},

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) = 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) = 1x + m, x = \(-\frac{m}{1}\)

(vii) p(x) = 3x^{2} – 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) = x^{2} – 1

Therefore, the value of polynomial p(x) at x = 1 is

p(1) = 1^{2} – 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) = x^{2} – 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) = x^{2}

Therefore, the value of polynomial p(x) at x = 0 is

p(0) = 0^{2} = 0

yes x = 0 is the zero of polynomial p(x) = x^{2}.

(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) = 3x^{2} – 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.