Jharkhand Board JAC Class 9 Maths Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.2

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Question 1.

Find the value of the polynomial 5x + 4x^{2} + 3 at

(i) x = 0

(ii) x = – 1

(iii) x = 2 ,

Answer:

(i) p(x) = 5x + 4x^{2} + 3

p(0) = 5(0) + 4(0)^{2} + 3 = 3

(ii) p(x) = 5x + 4x^{2} + 3

P(-1) = 5(-1) + 4(-1)^{2} + 3

= -5 + 4(1) + 3 = 2

(iii) p(x) = 5x + 4x^{2} + 3

p(2) = 5(2) + 4(2)^{2} + 3

= 10 + 16 + 3 = 29

Question 2.

Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y) = y^{2} – y + 1

(ii) p(x) = x^{3}

(iii) p(t) = 2 + t + 2t^{2} – t^{3}

(iv) p(x) = (x – 1) (x + 1)

Answer:

(i) p(y) = y^{2} – y + 1

P(0) = (0)^{2} – (0) + 1 = 1

p(1) = (1)^{2} – (1) + 1 = 1

p(2) = (2)^{2} – (2) + 1 = 3

(ii) p(x) = x^{3}

P(0) = (0)^{3} = 0

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

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

(iii) p(t) = 2 + t + 2t^{2} – t^{3}

p(0) = 2 + 0 + 2 (0)^{2} – (0)^{3} = 2

P(1) = 2 + (1) + 2(1)^{2} – (1)^{3}

= 2 + 1 + 2 – 1 = 4

p(2) = 2 + 2 + 2(2)^{2} – (2)^{3}

= 2 + 2 + 8 – 8 = 4

(iv) p(x) = (x – 1) (x + 1)

P(0) = (0 – 1) (0 + 1) = (-1) (1) = – 1

p(1) = (1 – 1) (1 + 1) = 0(2) = 0

p(2) = (2 – 1) (2 + 1) = 1(3) = 3

Page – 35

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 = \(\frac{1}{2}\)

Answer:

(i) If x = \(– \frac{1}{3}\) is a zero of polynomial p(x) then p(\(– \frac{1}{3}\)) = 0

p(\(– \frac{1}{3}\)) = 3(\(– \frac{1}{3}\)) + 1 = -1 + 1 = 0

Therefore, x = \(– \frac{1}{3}\) is a zero of polynomial p(x) = 3x + 1.

(ii) If x = \(\frac{4}{5}\) is a zero of polynomial

p(x) = 5x – π then p(\(\frac{4}{5}\)) should be 0.

p(\(\frac{4}{5}\)) = 5(\(\frac{4}{5}\)) – π = 4 – π ≠ 0

Therefore, x = \(\frac{4}{5}\) is not a zero of given polynomial p(x) = 5x – π.

(iii) If x = 1 and x = – 1 are zeroes of polynomial p(x) = x^{2} – 1, then p(1) and p(-1) each should be 0.

p(1) = (1)^{2} – 1 = 0 and

P(-1) = (-1)^{2} – 1 = 0

Hence, x = 1 and – 1 are zeroes of the polynomial p(x) = x^{2} – 1.

(iv) If x = -1 and x = 2 are zeroes of polynomial p(x) = (x + 1) (x – 2), then p( -1) and p(2) each should be 0.

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

p(2) = (2 + 1) (2 – 2) = 3 (0) = 0

Therefore, x = – 1 and x = 2 are zeroes of the polynomial p(x) = (x+1) (x – 2).

(v) If x = 0 is a zero of polynomial p(x) = x^{2}, then p(0) should be zero.

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

Hence, x = 0 is a zero of the polynomial p(x) = x^{2}.

(vi) If x = \(– \frac{m}{l}\) is a zero of polynomial p(x) = lx + m, then p(\(– \frac{m}{l}\))= 0 should be 0.

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

Therefore x = (\(– \frac{m}{l}\)) is a zero of polynomial p(x) = lx + m

(vii) If x = \(-\frac{1}{\sqrt{3}}\) and x = \( \frac{2}{\sqrt{3}}\) are zeros of polynomial p(x) = 3x^{2} – 1, then p(\(-\frac{1}{\sqrt{3}}\)) and p(\( \frac{2}{\sqrt{3}}\))should be 0.

p(\( \frac{-1}{\sqrt{3}}\)) = \(3\left(\frac{-1}{\sqrt{3}}\right)^2\) – 1 = 3(\(\frac{1}{3}\))-1 = 1 – 1 = 0

p(\( \frac{2}{\sqrt{3}}\)) = \(3\left(\frac{2}{\sqrt{3}}\right)^2\) – 1 = 3(\(\frac{4}{3}\)) – 1 = 4 – 1 = 3 ≠ 0

Therefore, x = \( \frac{-1}{\sqrt{3}}\) is a zero of polynomial p(x) = 3x^{2} – 1 but, x = \( \frac{2}{\sqrt{3}}\) is not a zero of this polynomial.

(viii) If x = \(\frac{1}{2}\) is a zero of polynomial p(x) = 2x + 1then p(\(\frac{1}{2}\)) should be 0.

p(\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\)) + 1 = 1 + 1 = 2 ≠ 0

Therefore, x = \( \frac{1}{2}\) is not a zero of given polynomial p(x) = 2x + 1

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, d, are real numbers.

Answer:

(i) p(x) = x + 5

p(x) = 0

x + 5 = 0

x = – 5

Therefore, x = – 5 is a zero of polynomial p(x) = x + 5 .

(ii) p(x) = x – 5

p(x) =0

x – 5 = 0

x = 5

Therefore, x = 5 is a zero of polynomial p(x) = x – 5.

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

p(x) = 0

2x + 5 = 0

2x = – 5

x = \(– \frac{5}{2}\)

Therefore, x = \(– \frac{5}{2}\) is a zero of polynomial p(x) = 2x + 5.

(iv) p(x) = 3x – 2

p(x) = 0

3x – 2 = 0

x = \( \frac{2}{3}\)

Therefore, x = \( \frac{2}{3}\) is a zero of polynomial p(x) = 3x – 2.

(v) p(x) = 3x

p(x) = 0

3x = 0

x = 0

Therefore, x = 0 is a zero of polynomial p(x) = 3x.

(vi) p(x) = ax, a ≠ 0

p(x) = 0

ax = 0

x = 0

Therefore, x = 0 is a zero of polynomial p(x) = ax.

(vii) p(x) = cx + d, c ≠ 0

p(x) = 0

cx + d = 0

x = \(– \frac{d}{c}\)

Therefore, x = \(– \frac{d}{c}\) is a zero of polynomial p(x) = cx + d.