HBSE 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.2

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 – 4x² + 3 at:
(i) x = 0
(ii) x = – 1
(iii) x = 2.
Solution :
Let p(x) = 5x – 4x² + 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) = y² – y + 1
(ii) p(t) = 2 + t + 2t² – t³
(iii) p(x) = x³
(iv) p(x) = (x – 1)(x + 1).
Solution :
(i) We have,
p(y) = y²y + 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 + 2t² – 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)= x³
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) = 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
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) = x² – 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)= x²
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) = 3x² – 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).