Haryana State Board HBSE 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 2 Polynomials Exercise 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) At x = 0, the value of the polynomial p(x) is given by

p(0) = 5 × 0 – 4 × (0)^{2} + 3

= 0 – 4 × 0 + 3

= 0 – 0 + 3 = 3

Hence, p(0) = 3.

(ii) At x = – 1, the value of the polynomial p(x) is given by

p(-1) = 5 × (-1) – 4 × (-1)^{2} + 3

= – 5-4 + 3 = -6 Hence, p(-1)=-6.

(iii) At x = 2, the value of the polynomial p(x) is given by

p(2) = 5 × 2 – 4 × (2)^{2} + 3

= 10 – 16 + 3 = – 3 Hence, p(2) = – 3.

Question 2.

Find p(o), 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) We have,

p(y) = y^{2}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.

Hence, p(0) = 1, p(1) = 1 and p(2) = 3.

(ii) We have, p(t) = 2 + t + 2t^{2} – 3

p(0) = 2 + 0 + 2 × (0)^{2} – (0)

= 2 + 0 + 0 – 0 = 2.

p(1) = 2 + 1 + 2 × (1) – (1)

= 2 + 1 + 2 – 1.4

and p(2) = 2 + 2 + 2 × (2) – (2)

= 2 + 2 + 8 – 8 = 4.

Hence, p(0) = 2, p(1) = 4 and p(2)=4.

(iii) We have,

p(x)= x^{3}

p(0) = (0)^{3} = 0.

p(1) = (1)^{3} = 1. and

p(2) = (2)^{3} = 8.

Hence, p(0) = 0; p(1)= 1 and p(2) = 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.

Hence, p(0) = – 1, p(1) = 0 and

p(2) = 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) We have,

p(x) = 3x + 1

The value of the polynomial p(x) at x = – \(\frac {1}{3}\) is given by

p(\(\frac {-1}{3}\)) = 3 × (\(\frac {-1}{3}\)) + 1

= -1 + 1 = 0

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

(ii) We have,

p(x) = 5x – π

The value of the polynomial p(x) at x = \(\frac {4}{5}\) is given by

p(\(\frac {4}{5}\)) = 5 × \(\frac {4}{5}\) – π = 4 – π

Hence, \(\frac {4}{5}\) is not zero of the polynomial p(x).

(iii) We have,

p(x) = x^{2} – 1

The value of the polynomial p(x) at x = 1 is given by

p(1) = (1)^{2} – 1 = 1 – 1 = 0

The value of the polynomial p(x) at x = -1 is given by

p(-1) = (-1)^{2} – 1 = 1 – 1 = 0

Hence, 1, – 1 both are the zeroes of the polynomial p(x).

(iv) We have,

p(x)=(x + 1)(x – 2)

The value of the polynomial p(x) at x = – 1 is given by

p(-1) = (-1 + 1)(-1 – 2)

= 0 × (-3) = 0

The value of the polynomial p(x) at x = 2 is given by

p(2) = (2 + 1) (2 – 2)

= 3 × 0 = 0

Hence, – 1 and 2 both are zeroes of the polynomial p(x).

(v) We have,

p(x)= x^{2}

The value of the polynomial p(x) at x = 0 is given by

p(0) = (0)^{2} = 0

Hence, 0 is zero of polynomial p(x).

(vi) We have,

p(x) = lx + m

The value of the polynomial p(x) at x = \(\frac {-m}{l}\) is given by

P(\(\frac {-m}{l}\)) = l × \(\frac {-m}{l}\) + m

= – m + m = 0

Hence, \(\frac {-m}{l}\) is zero of the polynomial p(x).

(vii) We have,

P(x) = 3x^{2} – 1

The value of the polynomial p(x) at x = \(\frac{-1}{\sqrt{3}}\) is given by

\(p\left(\frac{-1}{\sqrt{3}}\right)=3 \times\left(\frac{-1}{\sqrt{3}}\right)^2\) – 1

= 3 × \(\frac {1}{3}\) – 1 = 1 – 1 = 0

The value of the polynomial p(x) at x = \(\frac{2}{\sqrt{3}}\) is given by

\(p\left(\frac{2}{\sqrt{3}}\right)=3 \times\left(\frac{2}{\sqrt{3}}\right)^2\) – 1

= 3 × \(\frac {4}{3}\) – 1 = 4 – 1 = 3

Hence, \(\frac{-1}{\sqrt{3}}\) is a zero of the polynomial p(x) but \(\frac{2}{\sqrt{3}}\) is not.

(viii) We have,

p(x) = 2x + 1

The value of the polynomial p(x) at x = \(\frac {1}{2}\) is given by

p(\(\frac {1}{2}\)) = 2 × \(\frac {1}{2}\) + 1 = 1 + 1 = 2

Hence, \(\frac {1}{2}\) is not a zero of the polynomial 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; where c, d are real numbers.

Solution :

(i) We have,

p(x) = x + 5

For zero of the polynomial, p(x) = 0

⇒ 0 = x + 5

⇒ x = – 5

Hence, – 5 is the zero of the polynomial p(x).

(ii) We have p(x) = x – 5

For the zero of the polynomial, p(x) = 0

⇒ 0 = x – 5

⇒ x = 5

Hence, 5 is the zero of the polynomial p(x).

(iii) p(x) = 2x + 5

For the zero of the polynomial, p(x) = 0

⇒ 0 = 2x + 5

⇒ 2x = -5

⇒ x = \(\frac {-5}{2}\)

Hence, \(\frac {-5}{2}\) is the zero of the polynomial p(x).

(iv) We have,

p(x)= 3x – 2

For the zero of the polynomial, p(x) = 0

⇒ 0 = 3x – 2

⇒ 3x = 2

⇒ x = \(\frac {2}{3}\)

Hence, \(\frac {2}{3}\) is the zero of the polynomial p(x).

(v) We have, p(x) = 3x For the zero of the polynomial, p(x) = 0

⇒ 0 = 3x

⇒ x = 0

Hence, 0 is the zero of the polynomial p(x).

(vi) We have

p(x) = ax, a ≠ o

For the zero of the polynomial, p(x) = 0

⇒ 0 = ax

⇒ x = 0

Hence, is the zero of the polynomial p(x).

(vii) We have p(x) = cx + d, c ≠ 0; where c, d are real numbers

For the zero of the polynomial, p(x) = 0

⇒ 0= cx + d

⇒ cx = – d

⇒ x = \(\frac {-d}{c}\)

Hence, \(\frac {-d}{c}\) is the zero of the polynomial p(x).