AP 9th Class Maths 2nd Chapter Polynomials Exercise 2.2 Solutions

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 – 4x² + 3 at
i) x = 0
ii) x = -1
iii) x = 2
Solution:
i) x = 0
Given P(x) = 5x – 4x² + 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 – 4x² + 3 at x = 0 is 3.

ii) x = -1
Given P(x) = 5x – 4x² + 3, then
P(-1) = 5(-1) – 4(-1)² + 3
= -5 – 4 + 3
= – 9 + 3 = – 6
Value of P(x) = 5x – 4x² + 3 at x = – 1 is (-6).

iii) x = 2
Given P(x) = 5x – 4x² + 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 – 4x² + 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) = y² – y + i
ii) p(t) = 2 + t + 2t² -13
iii) P(x) = x³
iv) p(x) = (x – 1) (x + 1)
Solution:
i) p(y) = y² – y + 1, then
p(0) = 0² – 0 + 1
= 0 – 0 + 1 = 1
p(1) = 1² – 1 + 1
= 1 – 1 + 1
= 1
p(2) = 2² – 2 + 1
= 4 – 2 + 1 = 3

ii) p(t) = 2 + t + 2t² – 13
p(0) = 2 + 0 + 2(0)² – 0³ = 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) = x³
p(0) = 03 = 0 p(0) = 0
p(1) = 1³ = 1 × 1 × 1 = 1
p(1) = 1
p(2) = 2³ = 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) = x² – 1; x = 1, -1
(iv) p(x) = (x + 1) (x – 2); x = – 1, 2
(v) p(x) = x²; x = o
(vi) p(x) = lx + m ; x = \(\frac{-\mathbf{m}}{\mathbf{l}}\)
(vii) p(x) = 3x² – 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) = x² – 1; x = 1, -1
then p(1) = 1² – 1 = 1 – 1 = 0
∴ p(1) = 0
Hence x = 1 is a zero of p(x) = x² – 1.

p(-1) = (-1)² -1 = 1 – 1 = 0
∴ p(-1) = 0
Hence x = -1 is a zero of p(x) = x² – 1.
∴ 1, -1 are two zeros of p(x) = x² – 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) = x²; 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