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Question
Factorise:
x3 + x2 – 4x – 4
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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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