Factor $(a^2+2a)^2-2(a^2+2a)-3$ completely I have this question that asks to factor this expression completely:
$$(a^2+2a)^2-2(a^2+2a)-3$$
My working out:
$$a^4+4a^3+4a^2-2a^2-4a-3$$
$$=a^4+4a^3+2a^2-4a-3$$
$$=a^2(a^2+4a-2)-4a-3$$
I am stuck here. I don't how to proceed correctly.

Although your working is correct, you won't be able to proceed further. The trick is to let $y=a^2+2a$. Then your expression becomes
$y^2-2y-3$
$=(y-3)(y+1)$
$=(a^2+2a-3)(a^2+2a+1)$
$=(a+3)(a-1)(a+1)^2$
