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RBSE Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

April 14, 2022 by Prasanna Leave a Comment

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.2

Question 1.
Find the value of the polynomial 5x – 4x2 + 3 at:
(i) x = 0
Answer:
Given p(x) = 5x – 4x2 + 3
(i) At x = 0: p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3

(ii) x = – 1
Answer:
At x = – 1: p(- 1) = 5(- 1) – 4(- 1)2 + 3 = – 5 – 4 + 3 = – 6

(iii) x = 2
Answer:
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
Answer:
Given, 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) p(t) = 2 + t + 2t2 – t3
Answer:
Given: 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 – 1 = 5 – 1 = 4
and p(2) = 2 + 2 + 2(2)2 – (2)3 = 2 + 2 + 8- 8 = 4

(iii) p(x) = x3
Answer:
Given: p(x) = x3
p(0) = (0) = 0,
p(1) = (1) = 1
and p(2) = (2) = 8

(iv) p(x) = (x – 1) (x + 1)
Answer:
Given: 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

RBSE Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

Question 3.
Verify whether the following are zeroes of the poiynomial, indicated against them.
(i) p(x) = 3x + 1, x = -1
Answer:
Given: 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)

(ii) p(x) = 5x – π, x = \(\frac{4}{5}\)
Answer:
Given: p(x) = 5x – π
At x = \(\frac{4}{2}\)
p\(\left(\frac{4}{5}\right)\) = 5\(\left(\frac{4}{5}\right)\) – π = 4 – π ≠ 0
∴ \(\frac{4}{2}\) is not a zero of p(x)

(iii) p(x) = x2 – 1, x = 1, -1
Answer:
Given: p(x) = x2
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) p(x) = (x + 1)(x – 2), x = 1, 2
Answer:
Given: p(x) = (x + 1)(x – 2)
At x = -1, p(x) = (-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 a zero of p(x)

(v) p(x) = x2, x = 0
Answer:
Given: p(x) = x2
At x = 0, p(0) = (0)2 = 0
∴ 0 is a zero of p(x)

(vi) p(x) = lx + m, x = –\(\frac{m}{l}\)
Answer:
Given: p(x) = lx + m
At x = –\(\frac{m}{l}\), \(\left(-\frac{m}{l}\right)\) = l\(\left(-\frac{m}{l}\right)\) = -m + m = 0
∴ \(\left(\frac{-m}{l}\right)\) is a zero of x.

(vii) p(x) = 3x2 – 1, x = \(-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
Answer:
Given: p(x) = 3x2 – 1
At x = \(-\frac{1}{\sqrt{3}}\), p\(\left(-\frac{1}{\sqrt{3}}\right)\) = 3\(\left(\frac{1}{\sqrt{3}}\right)\)2 – 1 = 3 × \(\frac{1}{3}\) – 1 = 1 – 1 = 0
∴ \(-\frac{1}{\sqrt{3}}\) is a zero of x.

At x = \(\frac{2}{\sqrt{3}}\), p\(\left(\frac{2}{\sqrt{3}}\right)\) = 3\(\left(\frac{2}{\sqrt{3}}\right)\)2 – 1 = 3 × \(\frac{4}{3}\) – 1 = 4 – 1 = 3 ≠ 0
∴ \(\frac{2}{\sqrt{3}}\) is not a zero of x.

(viii) p(x) = 2x + 1, x = \(\frac{1}{2}\)
Answer:
Given: p(x) = 2x + 1
At x = \(\frac{1}{2}\), p\(\left(\frac{1}{2}\right)\) = 2\(\left(\frac{1}{2}\right)\) + 1 = 1 + 1 = 2 ≠ 0
∴ \(\frac{1}{2}\) is not a zero of x.

RBSE Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

Question 4.
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
Answer:
If p(x) = 0
or x + 5 = 0 ⇒ x = -5
∴ -5 is a zero of the polynomial x + 5.

(ii) p(x) = x – 5
Answer:
If p(x) = 0
or x – 5 = 0 ⇒ x = 5
∴ 5 is a zero of the polynomial x -5

(iii) p(x) = 2x + 5
Answer:
If p(x) = 0
or 2x + 5 = 0 ⇒ x = –\(\frac{5}{2}\)
∴ –\(\frac{5}{2}\) is a zero of the polynomial 2x + 5.

(iv) p(x) = 3x – 2
Answer:
If p(x) = 0
or 3x – 2 = 0 ⇒ x = \(\frac{2}{3}\)
∴ \(\frac{2}{3}\) is a zero of the polynomial 3x – 2.

(v) p(x) = 3x
Answer:
If p(x) = 0
or 3x = 0 ⇒ x = 0
∴ 0 is a zero of the polynomial 3x.

(vi) p(x) = ax, a ≠ 0
Answer:
If p(x) = ax, a ≠ 0
or ax = 0 or x = 0
∴ 0 is a zero of the polynomial ax.

(vii) p(x) = cx + d, c ≠ 0, e, d are real numbers.
Answer:
If p(x) = 0, c ≠ 0
or cx + d = 0 ⇒ x = –\(\frac{d}{c}\)
∴ –\(\frac{d}{c}\) is a zero of the polynomial cx + d.

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