Gujarat Board Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.2

03/06/2023 Ajit kumar 191 Views 0 Comments

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

x² – 2x – 8
4s² – 4s + 1
6x³ – 3 – 7x
4u² + 8u
t² – 15
3x² – x – 4

Solution:
1. p(x) = x² – 2x – 8
= x² – 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) = 4s² – 4s + 1
= 4s² – 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) = 6x² – 3 – 7x
p(x) = 6x² – 7x – 3
= 6x² – 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) = 4u² + 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. t² – 15
P(t) = t² – 15
For zeroes p(t) = 0
t² – 15 = 0

6. p(x) = 3x² – x – 4
p(x) = 3x² – 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