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Factorise: x3 + x2 – 4x – 4 - Mathematics

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Question

Factorise:

x3 + x2 – 4x – 4

Sum
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Solution

Let p(x) = x3 + x2 – 4x – 4

Constant term of p(x) = – 4

Factors of – 4 are ±1, ±2, ±4

By trial, we find that p(–1) = 0, so (x + 1) is a factor of p(x)

Now, we see that x3 + x2 – 4x – 4

= x2(x + 1) – 4(x + 1)

= (x + 1)(x2 – 4)  ...[Taking (x + 1) common factor]

Now, x2 – 4 = x2 – 22

= (x + 2)(x – 2)  ...[Using identity, a2 – b2 = (a – b)(a + b)]

∴ x3 + x2 – 4x – 4 = (x + 1)(x – 2)(x + 2)

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Chapter 2: Polynomials - Exercise 2.3 [Page 21]

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NCERT Exemplar Mathematics [English] Class 9
Chapter 2 Polynomials
Exercise 2.3 | Q 24. (iii) | Page 21
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