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.

If you assign $$ a^2 + 2a = x $$ you'll get: $$ x^2 - 2x - 3 $$
Considering that $$ x^2 - 2x - 3 = (x - 3)(x+1) $$ you'll get: $$ (a^2 + 2a - 3)(a^2 + 2a+1) = (a + 3)(a - 1)(a + 1)^2 $$ 
