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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 9

Factorise the following polynomials using synthetic division: x3 + x2 – 14x – 24 - Mathematics

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Question

Factorise the following polynomials using synthetic division:

x3 + x2 – 14x – 24

Sum

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)

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Synthetic Division
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Chapter 3: Algebra - Exercise 3.8 [Page 114]

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Samacheer Kalvi Mathematics [English] Class 9 TN Board
Chapter 3 Algebra
Exercise 3.8 | Q 1. (iv) | Page 114
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