Mathematics
Factorise :
(a2 - a)(4a2 - 4a - 5) - 6
Factorisation
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Answer
Given,
(a2 - a)(4a2 - 4a - 5) - 6
= (a2 - a)[4(a2 - a) - 5] - 6
Substituting a2 - a = x, we get :
= x(4x - 5) - 6
= 4x2 - 5x - 6
= 4x2 - 8x + 3x - 6
= 4x(x - 2) + 3(x - 2)
= (x - 2)(4x + 3)
= (a2 - a - 2)[4(a2 - a) + 3]
= (a2 - a - 2)(4a2 - 4a + 3)
= (a2 - 2a + a - 2)(4a2 - 4a + 3)
= [a(a - 2) + 1(a - 2)](4a2 - 4a + 3)
= (a - 2)(a + 1)(4a2 - 4a + 3).
Hence, (a2 - a)(4a2 - 4a - 5) - 6 = (a - 2)(a + 1)(4a2 - 4a + 3).
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