# Gujarat Board Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.2

Gujarat Board Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.2

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.2

Question 1.

Find the zeroes of the following polynomial and verify the relationship between the zeroes and the coefficients

- x
^{2}– 2x – 8 - 4s
^{2}– 4s + 1 - 6x
^{3}– 3 – 7x - 4u
^{2}+ 8u - t
^{2}– 15 - 3x
^{2}– x – 4

Solution:

1. p(x) = x^{2} – 2x – 8

= x^{2} – 4x + 2x – 8

[By splitting middle term]

= x(x – 4) + 2(x – 4)

= (x – 4) (x + 2)

For zeroes of p(x),

p(x) = 0

⇒ (x – 4) (x + 2) = 0

⇒ x – 4 = 0

or x + 2 = 0

⇒ x = 4

or x = -2

⇒ x = 4, – 2.

Zeroes of p(x) are 4 and -2.

⇒ α = 4, β = -2

2. p(s) = 4s^{2} – 4s + 1

= 4s^{2} – 2s – 2s + 1

[By splitting middle term]

= 2s(2s – 1) – 1 (2s – 1)

(2s – 1) (2s – 1)

For zeroes of p(s),

P(s) = 0

⇒ (2s – 1) (2s – 1) = 0

⇒ 2s – 1 = 0

or 2s – 1 = 0

3. p(x) = 6x^{2} – 3 – 7x

p(x) = 6x^{2} – 7x – 3

= 6x^{2} – 9x + 2x – 3

= 3x(2x – 3) + 1(2x – 3)

= (2x – 3) (3x + 1)

For zeroes p(x) = 0

⇒ (2x – 3) (3x + 1) = 0

⇒ 2x – 3 = 0

or 3x + 1 = 0

4. p(u) = 4u^{2} + 8 u

p(u) = 4u (u + 2)

For zeroes, p(u) = 0

⇒ 4 u(u + 2) = 0

⇒ u(u + 2) = 0

⇒ u = 0

or u + 2 = 0

⇒ u = 0

⇒ u = -2

So, zeroes of p(u) are 0 and -2.

⇒ α = 0, β = -2

Verification of relationship between zeroes and coefficients:

Sum of the zeroes,

α + β = 0 + (-2) = -2

5. t^{2} – 15

P(t) = t^{2} – 15

For zeroes p(t) = 0

t^{2} – 15 = 0

6. p(x) = 3x^{2} – x – 4

p(x) = 3x^{2} – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4)

= (3x – 4) (x + 1)

For zeroes, p(x) = 0

⇒ (3x – 4)(x + 1) = 0

⇒ 3x – 4 = 0

or x + 1 = 0

Question 2.

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

6. 4, 1

Sum of the zeroes, S = 4

Product of the zeroes, P = 1

∴ Required polynomial is

p(x) = K[x^{2} – Sx + P]

p(x) = x^{2} – 4x + 1

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