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 – 4x^{2} + 3 at:

(i) x = 0

Answer:

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

Answer:

Given, 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

(ii) p(t) = 2 + t + 2t^{2} – t^{3}

Answer:

Given: p(t) = 2 + t + 2t^{2} – t^{3}

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) = x^{3}

Answer:

Given: p(x) = x^{3}

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

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) = x^{2} – 1, x = 1, -1

Answer:

Given: p(x) = x^{2}

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) = x^{2}, x = 0

Answer:

Given: p(x) = x^{2}

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) = 3x^{2} – 1, x = \(-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)

Answer:

Given: p(x) = 3x^{2} – 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.

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.

## Leave a Reply