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 – 4x^{2} + 3 at

(i) x = 0

(ii) x = – 1

(iii) x = 2

Solution:

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

∴ p(0) = (0)^{2} – (0) + 1 = 1

p(y) = y^{2} – 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 + 2t^{2} – t^{3}

∴ 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} – 2^{3}

= 4 + 2 x 4 – 8

= 4 + 8 – 8 = 4

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

∴ 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) = 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) = lx+m, x = \(\frac { -m }{ l }\)

(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) 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) = x^{2} – 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) = x^{2}, 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)