**KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2** are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.2.

## Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.2

**KSEEB Solutions For Class 9 Maths Polynomials Question 1.**

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

i) x = 0

ii) x = -1

iii) x = 2

Answer:

i) f(x) = 5x – 4x^{2} + 3 x = 0 then,

f(0) = 5(0) – 4(0)^{2} + 3

=0 – 0 + 3

f(0) = 3

ii) f(x) = 5x – 4x^{2} + 3 x = -1 then,

f(-1) = 5(-1) – 4(-1)2 + 3

= -5 – 4(+1) + 3

= -5 – 4 + 3

= -9 + 3

f(-1) = -6

iii) f(x) = 5x – 4x^{2} + 3 x = 2 then,

f(2) = 5(2) – 4(2)^{2} + 3

= 5(2) – 4(4) + 3

= 10 – 16 + 3

= 13 – 16

f(2) = -3

**9th Maths Polynomials Exercise 4.2 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(t) = 2 + t + 2t^{2} – t^{3}

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

iv) p(x) = (x – 1) (x + 1)

Answer:

i) (a) p(y) = y^{2} – y + 1

p(0) = (0)^{2} – 0 + 1

= 0 – 0 + 1

∴ p(0) = =1

(b) p(y) = y^{2} – y + 1

p(1) = (1)^{2} – 1 + 1

= 1 – 1 + 1

∴ p(1) = 1

(c) p(y) = y^{2} – y +

p(2) = (2)^{2} – 2 + 1

= 4 – 2 + 1

= 5 – 2

∴ p(2) = 3

ii) (a) p(t) = 2 + t + 2t^{2} – t^{3}

p(0) = 2 + 0 + 2^{2} – (0)^{3}

= 2 + 0 + 0 – 0

∴ p(0) = 2

(b) p(t) = 2 + t + 2t^{2} – t^{3}

p(1) = 2 + 1 + 2(1)^{2} – (1)^{3}

= 2 + 1 + 2(1) – 1

= 2 + 1 + 2 – 1

∴ p(1) = 4

(c) p(t) – 2 + t + 2t^{2} – t^{3}

p(2) = 2 + 2 + 2(2)^{2} – (2)^{3}

= 2 + 2 + 2(4) – 8

= 2 + 2 + 8 – 8

∴ p(2) = 4

iii) (a) p(x) = x^{3}

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

p(0) = 0

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

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

∴ p(1) = 1

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

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

∴ p(2) = 8

iv) (a) p(x) = (x – 1)(x + 1)

p(x) = x^{2} – 1 [∵ (a + b)(a – b) = a^{2} – b]

p(0) = (0)^{2} – 1

= 0 – 1

∴ p(o) = -1

(b) p(x) = (x – 1)(x + 1)

p(x) = x^{2} – a [∵ (a + b)(a – b) = a^{2} – b^{2}]

p(1) = (1)^{2} – 1

= 1 – 1

∴ p(1) = 0

(c) p(x) = (x – 1)(x + 1)

p(x) = x^{2} – 1 [∵ (a + b)(a – b) = a^{2} – b^{2}]

p(2) = (2)^{2} – 1

= 4 – 1

∴ p(0) = 3

**KSEEB Solutions For Class 9 Maths Polynomials Exercise 4.2 Question 3.**

Verify whether the following are zeroes of the polynomial, induced 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) p(x) = 3x + 1; \(x=-\frac{1}{3}\)

Here value of polynomial is zero.

\(x=-\frac{1}{3}\) is not zero of the polynomial

ii) p(x) = 5x – π; \( x=\frac{4}{5}\)

Here value of polynomial is not zero.

\(x=\frac{4}{5}\) is not zero of the polynomial

iii) p(x) = x^{2} – 1; x = 1, -1

p(1) = (1)^{2} – 1

= 1 – 1

p(1) = 0

Here value of p(x) is zero.

hence its zero is 1.

p(x) = x^{2} – 1; x = -1

p(-1) = (-1)^{2} – 1

= 1 – 1

p(-1) = 0

Here value of p(x) is zero.

∴ -1 is zero.

iv) p(x) = (x – 1)(x – 2); x = -1, 2

p(x) = x^{2} – 2x + x – 2

p(x) = x^{2} – x + 2 x = -1

p(-1) = (-1)^{2} – (-1)^{2} + 2

= 1 + 1 + 2

p(-1) = 4

Here value of polynomila is not zero.

∴ -1 is not zero.

p(x) = x^{2} – x + 2 x = 2

p(2) = (2)^{2} – (2) + 2

= 4 – 2 + 2

p(-1) = 4

Here value of polynomial is not zero.

∴ 2 is not zero.

(v) p(x) = x^{2}; x = 0

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

p(0) = 0

Here value of p(x) is zero.

∴ 0 is its zero.

vi) p(x) = lx + m; \(x=-\frac{m}{l}\)

\(\mathrm{p}\left(-\frac{\mathrm{m}}{l}\right)=l\left(-\frac{\mathrm{m}}{l}\right)+\mathrm{m}\)

= -m + m

= 0

Here p(x) is zero.

∴ \(-\frac{\mathrm{m}}{l}\) is its zero.

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

viii) p(x) = 2x + 1; \(x=\frac{1}{2}\)

Here value of p(x) is not zero.

∴ \(\frac{1}{2}\) is not its zero.

**Polynomials Class 9 Exercise 4.2 Solutions 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, c, d are real numbers.

Answer:

i) p(x) = x + 5

Let p(x) =0, then,

p(x) = x + 5 = 0

x = 0 – 5

∴ x = -5

-5 is zero of p(x).

ii) p(x) = x – 5

If p(x) = 0, then

p(x) = x – 5 = 0

x = 0 + 5

∴ x = 5

5 is the zero of p(x).

iii) p(x) = 2x + 5

If p(x)= 0, then

p(x) = 2x + 5 = 0

2x = – 5

∴ \(x=\frac{-5}{2}\)

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

iv) p(x) = 3x – 2

If p(x)= 0, then

p(x) = 3x – 2 = 0

3x = 2

∴ \(x=\frac{2}{3}\)

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

v) p(x) = 3x

If p(x) = 0, then

p(x) = 3x = 0

∴ \(x=\frac{0}{3}\)

\(\frac{0}{3}\) is the zero of p(x)

vi) p(x) = ax, a ≠ 0

If p(x)= 0, then

p(x) = ax = 0

∴ \(x=\frac{0}{a}\) ∴ x = ∞(infinity)

∞ is the zero of p(x).

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

If p(x)= 0, then

p(x) = cx + d = 0

cx = 0 – d

cx = -d

∴ \(x=-\frac{d}{c}\)

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

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