Well-designed AP 9th Class Maths Textbook Solutions Chapter 2 Number System Exercise 2.2 offers step-by-step explanations to help students understand problem-solving strategies.

## Polynomials Class 9 Exercise 2.2 Solutions – 9th Class Maths 2.2 Exercise Solutions

Question 1.

Find the value of the polynomial 5x – 4x^{2} + 3 at

i) x = 0

ii) x = -1

iii) x = 2

Solution:

i) x = 0

Given P(x) = 5x – 4x^{2} + 3

Then the value of polynomial P(x) at x = 0 is

P(0) = 5(0) – 4(0)^{2} + 3

= 0 – 0 + 3 = 3.

Value of P(x) = 5x – 4x^{2} + 3 at x = 0 is 3.

ii) x = -1

Given P(x) = 5x – 4x^{2} + 3, then

P(-1) = 5(-1) – 4(-1)^{2} + 3

= -5 – 4 + 3

= – 9 + 3 = – 6

Value of P(x) = 5x – 4x^{2} + 3 at x = – 1 is (-6).

iii) x = 2

Given P(x) = 5x – 4x^{2} + 3, then

P(2) = 5(2) – 4(2)^{2} + 3

= 10 – 4(4) + 3

= 10 – 16 + 3

= -3

then the value of P(x) = 5x – 4x^{2} + 3 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 + i

ii) p(t) = 2 + t + 2t^{2} -13

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

iv) p(x) = (x – 1) (x + 1)

Solution:

i) p(y) = y^{2} – y + 1, then

p(0) = 0^{2} – 0 + 1

= 0 – 0 + 1 = 1

p(1) = 1^{2} – 1 + 1

= 1 – 1 + 1

= 1

p(2) = 2^{2} – 2 + 1

= 4 – 2 + 1 = 3

ii) p(t) = 2 + t + 2t^{2} – 13

p(0) = 2 + 0 + 2(0)^{2} – 0^{3} = 2

p(0) = 2

p(1) = 2 + 1 + 2(1)^{2} – 13

= 2 + 1 + 2(1) – 1

= 2 + 1 + 2 – 1 = 4

.-.p(1) = 4

p(2) = 2 + 2 + 2(2)2 – 23

= 2 + 2 + 2(4)-8

= 2 + 2+ 0 – 0 =4

p(2) = 4

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

p(0) = 03 = 0 p(0) = 0

p(1) = 1^{3} = 1 × 1 × 1 = 1

p(1) = 1

p(2) = 2^{3} = 2 × 2 × 2 = 8 ⇒ p(2) = 8

Question 3.

Verify whether the values of x given in each case are the zeroes of the polynomial or not ?

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

(ii) p(x) = 5x – 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 = o

(vi) p(x) = lx + m ; x = \(\frac{-\mathbf{m}}{\mathbf{l}}\)

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

(viii) f(x) = 2x – 1; x = \(\frac{1}{2}\)

Solution:

(i) p(x) at x = \(\frac{-1}{3}\) is

Img 1

∴ p(\(\frac{-1}{3}\)) = 0

Hence p(x) = \(\frac{-1}{3}\) is a zero of p(x).

(ii) p(x) = 5x – π ; x = \(\frac{4}{5}\)

Img 2

Hence x = \(\frac{4}{5}\) is not a zero of given p(x) = 5x – π.

(iii) p(x) = x^{2} – 1; x = 1, -1

then p(1) = 1^{2} – 1 = 1 – 1 = 0

∴ p(1) = 0

Hence x = 1 is a zero of p(x) = x^{2} – 1.

p(-1) = (-1)^{2} -1 = 1 – 1 = 0

∴ p(-1) = 0

Hence x = -1 is a zero of p(x) = x^{2} – 1.

∴ 1, -1 are two zeros of p(x) = x^{2} – 1.

(iv) p(x) = (x + 1) (x – 2); x = – 1, 2

p(-1) = (-1 + 1) (-1 – 2) = 0(-3) = 0

⇒ p(-1) = 0

and p(2) = (2 + 1) (2 – 2) = 3(0) = 0

⇒ p(2) = 0

So p(-1) = 0 and p(2) = 0 for

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

Hence -1 and 2 are two zeros of p(x) = (x + 1) (x – 2)

(v) p(x) = x^{2}; x = o

⇒ p(0) = 0² = 0

⇒ p(0) = 0

∴ ‘0’ is a zero of p(x) = x²

vi) p(x) = lx + m ; x = \(\frac{-\mathbf{m}}{\mathbf{l}}\)

Img 3

vii) p(x) = 3x² – 1; x = \(\frac{-1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)

Now,

Img 4

and for x = \(\frac{2}{\sqrt{3}}\)

p(x) = 3x² – 1

Img 5

Hence \(\frac{2}{\sqrt{3}}\) is not a zero of a p(x) = 3x² – 1.

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

Img 6

p(\(\frac{1}{2}\)) = 2 which is not equal to zero.

Hence \(\frac{1}{2}\) is not a zero of p(x) = 2x + 1.

Question 4.

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

i) (x) = x + 5

Solution:

To find zero of p(x), we put p(x) = 0

∴ p(x) = x + 5 = 0 ⇒ x = 0 – 5 = -5

∴ x = -5 is a zero of p(x) = x + 5.

Verification:

p(x) = x + 5

then p(-5) = -5 + 5 = 0

Hence -5 is a zero of p(x) = x + 5.

ii) p(x) = x – 5

Solution:

To find zero of p(x), we put p(x) = 0

So, p(x) = x + 5 ⇒ 0 = x = 0 + 5 = 5

So, -5 is a zero of p(x) = x – 5.

Verification:

p(x) = x – 5 ⇒ p(5) = 5 – 5 = 0

So, 5 is a zero of p(x).

iii) p(x) = 2x + 5

Solution:

p(x) = 2x + 5

then p(x) = 2x + 5 = 0 ⇒ 2x = 0 – 5 = -5

2x = -5 ⇒ x = \(\frac{-5}{2}\)

then (\(\frac{-5}{2}\)) is a zero of given,

p(x) = 2x + 5

Verification:

p(x) = 2x + 5

Img 7

Hence \(\frac{-5}{2}\) is a zero of given p(x) = 2x + 5.

iv) p(x) = 3x – 2

Solution:

p(x) = 3x – 2

p(x) = 3x – 2 = 0

(∵ for a zero p(x) = 0)

⇒ 3x = 2

⇒ x = \(\frac{2}{3}\)

∴ \(\frac{2}{3}\) is a zero of p(x) = 3x – 2.

v) p(x) = 3x

Solution:

p(x) = 3x

⇒ p(x) = 3x = 0 ⇒ x = \(\frac{0}{3}\) = 0

∴ ‘0’ is a zero of p(x) = 3x.

Verification:

p(x) = 3x .

then p(o) = 3(0) = 0

∴ ‘0’ is a zero of given p(x) = 3x

vi) p(x) = ax, a ≠ 0

then p(x) = ax = 0 ⇒ x = \(\frac{0}{3}\) = 0

∴ ‘0’ is a zero of given p(x) = ax, a ≠ 0

Verification:

p(x) = ax

then p(0) = a(0) = 0

:. ‘0’ is a zero of given p(x) = ax