Question
Factorise the following:$\ (t^2 - t)(4t^2 - 4t - 5) - 6$
✓
Answer
$(t^2 - t)(4t^2 - 4t - 5) - 6$
$= (t^2 - t)[4(t^2 - t) - 5] - 6$
$= a[4a - 5] - 6[$Taking $(t^2 - t) = a]$
$= 4a^2 - 5a - 6$
$= 4a^2 - 8a + 3a - 6$
$= 4a(a - 2) + 3(a - 2)$
$= (a - 2)(4a + 3)$
$= (t^2 - t - 2)[4(t^2 - t) + 3]$
$= (t^2 - 2t + t - 2)(4t^2 - 4t + 3)$
$= [t(t - 2) + 1(t - 2)](4t^2 - 4t + 3)$
$= [(t - 2)(t + 1)](4t^2 - 4t + 3)$
$= (t + 1)(t - 2)(4t^2 - 4t + 3).$
$= (t^2 - t)[4(t^2 - t) - 5] - 6$
$= a[4a - 5] - 6[$Taking $(t^2 - t) = a]$
$= 4a^2 - 5a - 6$
$= 4a^2 - 8a + 3a - 6$
$= 4a(a - 2) + 3(a - 2)$
$= (a - 2)(4a + 3)$
$= (t^2 - t - 2)[4(t^2 - t) + 3]$
$= (t^2 - 2t + t - 2)(4t^2 - 4t + 3)$
$= [t(t - 2) + 1(t - 2)](4t^2 - 4t + 3)$
$= [(t - 2)(t + 1)](4t^2 - 4t + 3)$
$= (t + 1)(t - 2)(4t^2 - 4t + 3).$
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