Mathematics
Factorise the following:
(x2 - x)(4x2 - 4x - 5) - 6
Factorisation
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Answer
(x2 - x)(4x2 - 4x - 5) - 6 = (x2 - x)[4(x2 - x) - 5] - 6
Let x2 - x = p.
(x2 - x)[4(x2 - x) - 5] - 6 = p(4p - 5) - 6
= 4p2 - 5p - 6
= 4p2 - 8p + 3p - 6
= 4p(p - 2) + 3(p - 2)
= (p - 2)(4p + 3)
= (x2 - x - 2)[4(x2 - x) + 3]
= (x2 - x - 2)(4x2 - 4x + 3)
= [x2 - 2x + x - 2](4x2 - 4x + 3)
= [x(x - 2) + 1(x - 2)](4x2 - 4x + 3)
= (x - 2)(x + 1)(4x2 - 4x + 3).
Hence, (x2 - x)(4x2 - 4x - 5) - 6 = (x - 2)(x + 1)(4x2 - 4x + 3).
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