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Question
Factorise the following:
(t2 - t)(4t2 - 4t - 5) - 6
Sum
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Solution
(t2 - t)(4t2 - 4t - 5) - 6
= (t2 - t)[4(t2 - t) - 5] - 6
= a[4a - 5] - 6 [Taking (t2 - t) = a]
= 4a2 - 5a - 6
= 4a2 - 8a + 3a - 6
= 4a(a - 2) + 3(a - 2)
= (a - 2)(4a + 3)
= (t2 - t - 2)[4(t2 - t) + 3]
= (t2 - 2t + t - 2)(4t2 - 4t + 3)
= [t(t - 2) + 1(t - 2)](4t2 - 4t + 3)
= [(t - 2)(t + 1)](4t2 - 4t + 3)
= (t + 1)(t - 2)(4t2 - 4t + 3).
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