GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.2

Gujarat Board GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.2

Question 1.
Find the value of the polynomial 5x – 4x² + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Solution:
Let p(x) = 5x – 4x² + 3
(i) Substituting x = 0 in p(x)
p(0) = 5 x (0) – 4 x (0)² + 3
p(0) = 3

(ii) Substituting x = – 1 in p(x)
p(- 1) = 5 x (- 1) – 4(- 1)² + 3
= – 5 – 4 + 3 = – 9 + 3 = – 6

(iii) Substituting x = 2 in p(x)
p(2) = 5 x 2 – 4(2)² + 3
= 10 – 4 x 4 + 3 = 13 – 16 = – 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) p(y) = y² – y + 1
∴ p(0) = (0)² – (0) + 1 = 1
p(y) = y² – y + 1
p(1) = (1)² – (1) + 1 = 1 – 1 = 1
p(2) = (2)² – 2 + 1 = 4 – 2 + 1 = 2 + 1 = 3

(ii) p(t) = 2 + t + 2t² – t³
∴ p(0) = 2 + (0) + 2(0)² – (0)³ = 2
p(1) = 2 + (1) + 2(1)² – (1)³
= 2 + 1 + 2 – 1 = 4
p(2) = 2 + 2 + 2(2)² – 2³
= 4 + 2 x 4 – 8
= 4 + 8 – 8 = 4

(iii) p(x) = x³
∴ p(0) = (0)³ = 0
p(1) = (1)³ = 1
p(2) = (2)³ = 8

(iv) p(x) = (x – 1) (x + 1)
∴ p(0) = (0 – 1) (0 + 1) = (- 1) x (1) = – 1
p(1) = (1 – 1) (1 + 1) = 0 x 2 = 0
p(2) = (2 – 1) (2 + 1) = 1 x 3 = 3

Question 3.
Verify whether the following are zeroes of the polynomial, indicated against them.
(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 = 0
(vi) p(x) = lx+m, x = \(\frac { -m }{ l }\)
(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) p(x) = 3x + 1, x = – \(\frac { 1 }{ 3 }\)
∴ p( \(\frac { -1 }{ 3 }\) ) = 3 x( \(\frac { – 1 }{ 3 }\) )
⇒ p( \(\frac { -1 }{ 3 }\) ) = – 1 + 1 = 0
Hence, x = ( \(\frac { -1 }{ 3 }\) ) is zero of p(x).

(ii) p(x) = 5x – π, x = \(\frac { 4 }{ 5 }\)
∴P( \(\frac { 4 }{ 5 }\) ) = 5 x ( \(\frac { 4 }{ 5 }\) ) – π = 4 – π ( 4 – π ≠ 0)
∴ x = \(\frac { 4 }{ 5 }\) is not a zero of p(x).

(iii) p(x) = x² – 1, x = 1, – 1
p(1) = (1)² – 1, 1 = 1 – 1 = 0
p(- 1) = (- 1)² – 1 = 1 – 1 = 0
∴ x = 1, and x = – 1 are zeroes

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
p(- 1) = (- 1 + 1) (- 1 – 2) .
= 0 x (- 3) = 0
p(2) = (2 + 1) (2 – 2) = 3 x (0)
∴ p(2) = 0
Hence x = – 1, and x = 2 are zeroes of p(x).

(v) p(x) = x², x = 0
p(0) = (0)² = 0
∴ x = 0 is a zero of p(x)

(vi) p(x) = lx + m, x = \(\frac { – m }{ l }\)
IMGG
∴ x = \(\frac{-1}{\sqrt{3}}\) is a zero of p(x)
but x = \(\frac{2}{\sqrt{3}}\) is not a zero of p(x).

(viii) p(x) = 2x + 1, x = \(\frac { 1 }{ 2 }\)
p( \(\frac { 1 }{ 2 }\) ) = 2( \(\frac { 1 }{ 2 }\) )+ 1 = 1 + 1 = 2 (2 ≠ 0)
∴ x = \(\frac { 1 }{ 2 }\) is not a zero of 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) p(x) = x + 5
Let p(x) = 0
⇒ x + 5 = 0
⇒ x = – 5
∴ x = – 5 is zero of polynomial p(x)

(ii) p(x) = x – 5
Let p(x) = 0
⇒ x – 5 = 0
⇒ x = 5
∴ x = 5 is zero of polynomial p(x)

(iii) p(x) = 2x + 5
Let p(x) = 0
⇒ 2x + 5 = 0
⇒ 2x = 5
∴ x = \(\frac { -5 }{ 2 }\) is zero of polynomial p(x)

(iv) p(x) = 3x – 2
Let p(x) = 0
⇒ 3x – 2 = 0
⇒ 3x = 2
∴ x = \(\frac { 2 }{ 3 }\) is zero of polynomial p(x)

(v) p(x) = 3x
Let p(x) = 0
⇒ 3x = 0
⇒ x = 0
∴ x = 0 is zero of polynomial p(x)

(vi) p(x) = ax, a ≠ 0
Let p(x) = 0
⇒ ax = 0
⇒ x = \(\frac { 0 }{ a }\)
⇒ x = 0
∴ x = 0 is zero of polynomial p(x)

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Let p(x) = 0
⇒ cx + d = 0
⇒ cx = – d
⇒ x = \(\frac { -d }{ c }\)
∴ x = \(\frac { -d }{ c }\) is zero of polynomial p(x)