Question
Factorise the following:$\ (p^2 + p)^2 - 8(p^2 + p) + 12$
✓
Answer
$(p^2 + p)^2 - 8(p^2 + p) + 12$
$= (p^2 + p)^2 - 6(p^2 + p) - 2(p^2 + p) + 12$
$= (p^2 + p)(p^2 + p - 6) - 2(p^2 + p - 6)$
$= (p^2 + p - 6)(p^2 + p - 2)$
$= (p^2+ 3p - 2p - 6)(p^2 + 2p - p - 2)$
$= [p(p + 3) - 2(p + 3)][p(p + 2) - 1(p + 2)]$
$= [(p + 3)(p - 2)][(p + 2)(p - 1)]$
$= (p + 3)(p - 2)(p + 2)(p - 1).$
$= (p^2 + p)^2 - 6(p^2 + p) - 2(p^2 + p) + 12$
$= (p^2 + p)(p^2 + p - 6) - 2(p^2 + p - 6)$
$= (p^2 + p - 6)(p^2 + p - 2)$
$= (p^2+ 3p - 2p - 6)(p^2 + 2p - p - 2)$
$= [p(p + 3) - 2(p + 3)][p(p + 2) - 1(p + 2)]$
$= [(p + 3)(p - 2)][(p + 2)(p - 1)]$
$= (p + 3)(p - 2)(p + 2)(p - 1).$
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