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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) x^{3} – 3x² – 10x + 24

Solution:

p(x) – x^{3} – 3x² – 10x + 24

p(1) = 1^{3} – 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) = 2^{3} – 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 x^{3} – 3x² – 10x + 24 = (x – 2) (x – 4) (x + 3)

(ii) 2x^{3} – 3x² – 3x + 2

Solution:

p(x) = 2x^{3} – 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 2x^{3} – 3x² – 3x + 2 = (x + 1) (x – 2) (2x – 1)

(iii) – 7x + 3 + 4x^{3}

Solution:

p(x) = – 7x + 3 + 4x^{3}

= 4x^{3} – 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 + 4x^{3} = (x – 1) (2x + 3) (2x – 1)

(iv) x^{3} + x² – 14x – 24

Solution:

p(x) = x^{3} + 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.

x^{3} + x^{2} – 14x – 24 = (x + 2) (x + 3) (x – 4)

(v) x^{3} – 7x + 6

Solution:

p(x) = x^{3} – 7x + 6

P( 1) = 1^{3} – 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.

∴ x^{3} – 7x + 6 = (x – 1) (x – 2) (x + 3)

(vi) x^{3} – 10x² – x + 10

p(x) = x^{3} – 10x^{2} – x + 10

= 1 – 10 – 1 + 10

= 11 – 11

= 0

∴ x – 1 is a factor

x^{2} – 9x – 10 = x^{2} – 10x + x – 10

= x(x – 10) + 1 (x – 10)

= (x – 10) (x + 1)

This (x – 1) (x + 1) (x – 10) are the factors.

∴ x^{3} – 10x^{2} – x + 10 = (x – 1) (x – 10) (x + 1)