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