Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.8

Question 1.
Factorise each of the following polynomials using synthetic division:
(i) x³ – 3x² – 10x + 24
Solution:
p(x) – x³ – 3x² – 10x + 24
p(1) = 1³ – 3(1)² – 10(1) + 24
= 1 – 3 – 10 + 24
= 25 – 13
≠ 0
x – 1 is not a factor

p(-1) = (-1)³ – 3(-1)² – 10(-1) + 24
= – 1 – 3(1) + 10 + 24
= -1 – 3 + 10 + 24
= 34 – 4
= 30
≠ 0
x + 1 is not a factor

p(2) = 2³ – 3(2)² – 10(2) + 24
= 8 – 3(4) – 20 + 24
= 8 – 12 – 20 + 24
= 32 – 32
= 0
∴ x – 2 is a factor

x² – x – 12 = x² – 4x + 3x – 12

= x(x – 4) + 3 (x – 4)
= (x – 4) (x + 3)
∴ The factors of x³ – 3x² – 10x + 24 = (x – 2) (x – 4) (x + 3)

(ii) 2x³ – 3x² – 3x + 2
Solution:
p(x) = 2x³ – 3x² – 3x + 2
P(1) = 2(1)³ – 3(1)² – 3(1) + 2
= 2 – 3 – 3 + 2
= 2 – 6
= -4
≠ 0
x – 1 is not a factor

P(-1) = 2(-1)³ – 3(-1)² – 3(-1) + 2
= -2 – 3 + 3 + 2
= 5 – 5
= 0
∴ x + 1 is a factor

2x² – 5x + 2 = 2x² – 4x – x + 2

= 2x(x – 2) – 1 (x – 2)
= (x – 2) (2x – 1)
∴ The factors of 2x³ – 3x² – 3x + 2 = (x + 1) (x – 2) (2x – 1)

(iii) – 7x + 3 + 4x³
Solution:
p(x) = – 7x + 3 + 4x³
= 4x³ – 7x + 3
P(1) = 4(1)³ – 7(1) + 3
4 – 7 + 3
= 7 – 7
= 0
∴ x – 1 is a factor

4x² + 4x – 3 = 4x² + 6x – 2x – 3

= 2x(2x + 3) – 1 (2x + 3)
= (2x + 3) (2x – 1)
∴ The factors of – 7x + 3 + 4x³ = (x – 1) (2x + 3) (2x – 1)

(iv) x³ + x² – 14x – 24
Solution:
p(x) = x³ + x² – 14x – 24
p(1) = (1)³ + (1)² – 14 (1) – 24
= 1 + 1 – 14 – 24
= -36
≠ 0
x + 1 is not a factor.

p(-1) = (-1)³ + (-1)² – 14(-1) – 24
= -1 + 1 + 14 – 24
= 15 – 25
≠ 0
x – 1 is not a factor.

p(2) = (-2)³ + (-2)² – 14 (-2) – 24
= -8 + 4 + 28 – 24
= 32 – 32
= 0
∴ x + 2 is a factor

x² – x – 12 = x² – 4x + 3x – 12

= x(x – 4) + 3 (x – 4)
= (x – 4) (x + 3)
This (x + 2) (x + 3) (x – 4) are the factors.
x³ + x² – 14x – 24 = (x + 2) (x + 3) (x – 4)

(v) x³ – 7x + 6
Solution:
p(x) = x³ – 7x + 6
P( 1) = 1³ – 7(1) + 6
= 1 – 7 + 6
= 7 – 7
= 0
∴ x – 1 is a factor

x² + x – 6 = x² + 3x – 2x – 6

= x(x + 3) – 2 (x + 3)
= (x + 3) (x – 2)
This (x – 1) (x – 2) (x + 3) are factors.
∴ x³ – 7x + 6 = (x – 1) (x – 2) (x + 3)

(vi) x³ – 10x² – x + 10
p(x) = x³ – 10x² – x + 10
= 1 – 10 – 1 + 10
= 11 – 11
= 0
∴ x – 1 is a factor

x² – 9x – 10 = x² – 10x + x – 10

= x(x – 10) + 1 (x – 10)
= (x – 10) (x + 1)
This (x – 1) (x + 1) (x – 10) are the factors.
∴ x³ – 10x² – x + 10 = (x – 1) (x – 10) (x + 1)