Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

Students can download Maths Chapter 3 Algebra Ex 3.8 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 1
x² – x – 12 = x² – 4x + 3x – 12
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 2
= x(x – 4) + 3 (x – 4)
= (x – 4) (x + 3)
∴ The factors of x3 – 3x² – 10x + 24 = (x – 2) (x – 4) (x + 3)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

(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
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 3
2x² – 5x + 2 = 2x² – 4x – x + 2
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 4
= 2x(x – 2) – 1 (x – 2)
= (x – 2) (2x – 1)
∴ The factors of 2x3 – 3x² – 3x + 2 = (x + 1) (x – 2) (2x – 1)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

(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
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 5
4x² + 4x – 3 = 4x² + 6x – 2x – 3
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 6
= 2x(2x + 3) – 1 (2x + 3)
= (2x + 3) (2x – 1)
∴ The factors of – 7x + 3 + 4x3 = (x – 1) (2x + 3) (2x – 1)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

(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
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 7
x² – x – 12 = x² – 4x + 3x – 12
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 8
= 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)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

(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
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 9
x² + x – 6 = x² + 3x – 2x – 6
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 10
= 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)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

(vi) x3 – 10x² – x + 10
p(x) = x3 – 10x2 – x + 10
= 1 – 10 – 1 + 10
= 11 – 11
= 0
∴ x – 1 is a factor
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 11
x2 – 9x – 10 = x2 – 10x + x – 10
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 12
= 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)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

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