Question

x(x^2-1)(x+2)-8 factorise

Answer 1

x(2x-1)(x+2)-8=x{2x(x+2)-1(x+2)}-8=x{2x²+3x-2}-8=2x³+3x²-8

Answer 2

answer is (x² + x + 2)(x² + x - 4)

x(x² - 1)(x + 2) - 8

= x(x - 1)[(x + 1)(x + 2)] - 8

= (x² - x)(x² + 3x + 2) - 8

= (x² + x - 2x)(x² + x + 2x + 2) - 8

let t = x² + x

= (t - 2x)(t + 2x + 2) - 8

= (t² + (2x + 2 - 2)t - 4x² - 4x - 8

= t² + 2t - 4(x² + x) - 8

= t² + 2t - 4t - 8

= t² - 2t - 8

= t² - 4t + 2t - 8

= t(t - 4) + 2(t - 4)

= (t + 2)(t - 4)

now putting t = (x² + x)

= (x² + x + 2)(x² + x - 4)

also read similar questions: 1/x-3+2/x-2=8/x

factorise this

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Factorise : x 2 + x/4 - 1/8

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