Factoring Expressions

Question

I can't seem to factor this expression:

$$2(2x^2-x)^2-3(2x^2-x)-9$$

So far, this is what I have done: $$(2x^2-x)(4x^2-2x-3)-9$$

I'm not sure what to do after this though, any hints?

Answer 1

hint :put $(2x^2-x) =t$

$2t^2-3t-9$

$2t^2-6t+3t-9$

$2t(t-3)+3(t-3)$

$(2t+3)(t-3)$

$[2(2x^2-x)+3][(2x^2-x)-3]$

Answer 2

The kicker, here, is that your expression is quadratic in form. Making the substitution $u=2x^2-x,$ we have the expression $$2u^2-3u-9,$$ which is easily factorable. Once we've factored that, we resubstitute $u=2x^2-x,$ and factor further as possible.