Bihar Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.2 Textbook Questions and Answers.

## BSEB Bihar Board Class 9th Maths Solutions 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) At x = 0: p(0) = 5(0) – 4(0)² + 3

= 0 – 0 + 3 = 3

(ii) At x = – 1 : p(- 1) = 5(- 1) – 4(- 1)² + 3

= – 5 – 4 + 3 = – 6.

(iii) At x = 2 : p(2) = 5(2) – 4(2)² + 3

= 10 – 16 + 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)(» + 1)

Solution:

(i) We have, p(y) = y² – y + 1

p(0) = (0)² – 0 + 1 = 0 – 0 + 1 = 1

p(1) = (1)² – 1 + 1 = 1 – 1 + 1 = 1,

and p(2) = (2)² – 2 + 1 = 4 – 2 + 1 = 3

(ii) We have, p(t) = 2 + t + 2t² – t³

∴ p(0) = 2 + 0 + 2(0)² – (0)³

= 2 + 0 + 0 – 0 = 2,

p(1) = 2 + 1 + 2(1)² – (1)³

= 2 + 1 + 2 – 3 = 5 – 3 = 2,

and p(2) = 2 + 2 + 2(2)² – (2)³

= 2 + 2 + 8 – 8 = 4

(iii) We have, p(x) = x³

∴ p( 0) = (0)³ = 0,

p(1) = (1)³ = 1,

and p(2) = (2)³ = 8

(iv) We have, p(x) = (x – 1)(x + 1)

p(0) = (0 – 1)(0 + 1) = (- 1)(1) = – 1,

p(1) = (1 – 1)(1 + 1) = (0)(2) = 0

and p(2) = (2 – 1)(2 + 1) = (1)(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) We have, p(x) = 3x + 1

At x = – \(\frac { 1 }{ 3 }\), P(-\(\frac { 1 }{ 3 }\)) = 3(-\(\frac { 1 }{ 3 }\)) + 1 = – 1 + 1 = 0

∴ – \(\frac { 1 }{ 3 }\) is a zero of p(x).

(i) We have, p(x) = 5x – π

At x = – \(\frac { 4 }{ 5 }\), P(\(\frac { 4 }{ 5 }\)) = 5(\(\frac { 4 }{ 5 }\)) – 4 = 4 – 4 = 0

∴ \(\frac { 4 }{ 5 }\) is a zero of p(x).

(iii) We have, p(x) = x³ – 1

At x = 1, p(1) = (1)² – 1 = 1 – 1 = 0

∴ 1 is a zero of p(x).

Also, at x = – 1 p(- 1) = (- 1)² – 1 = 1 – 1 = 0

∴ – 1 is a zero of p(x).

(iv) We have, p(x) = (x + 1)(x – 2)

At x = – 1, p(- 1) = (- 1 + 1)(- 1 – 2) = (0)(- 3) = 0

∴ – 1 is a zero of p(x).

Also, at x = 2, p(2) = (2 + 1)(2 – 2) = (3)(0) = 0

∴ 2 is zero of p(x).

(v) We have, p(x) = x²

At x = 0, p(0) = (0)² = 0

∴ 0 is a zero of p(x).

(vi) We have p(x) = lx + m

At x = – \(\frac { m }{ l }\), p(-\(\frac { m }{ l }\)) = l(\(\frac { -m }{ l }\)) + m = – m + m = 0

∴ \(\frac { -m }{ l }\) is a zero of p(x)

(viii) We have, p(x) = 2x + 1

At x = \(\frac { 1 }{ 2 }\), p(\(\frac { 1 }{ 2 }\)) = 2(\(\frac { 1 }{ 2 }\)) + 1 = 1 + 1 = 2.

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(xr) = 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 to solve p(x) = 0

⇒ x + 5 = 0 ⇒ x = – 5

∴ – 5 is a zero of the polynomial x + 5.

(ii) We have to solve p(x) = 0

⇒ x – 5 = 0 ⇒ x = 5

∴ 5 is a zero of the polynomial x – 5.

(iii) We have to solve p(x) = 0

2x + 5 = 0 ⇒ x = – \(\frac { 5 }{ 2 }\)

∴ – \(\frac { 5 }{ 2 }\) is a zero of the polynomial 2x + 5.

(iv) We have to solve p(x) = 0

3x – 2 – 0 ⇒ x = \(\frac { 2 }{ 3 }\)

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

(v) We have to solve p(x) = 0

⇒ 3x = 0 ⇒ x = 0

∴ 0 is a zero of the polynomial 3x.

(vi) We have to solve p(x) = ax, a ≠ 0

⇒ ax = 0 ⇒ x = 0

∴ 0 is a zero of the polynomial ax.

(vii) We have to solve p(x) = 0, c ≠ 0

⇒ cx + d = 0 ⇒ x = – \(\frac { d }{ c }\)

∴ – \(\frac { d }{ c }\) is a zero of the polynomial cx + d.