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.

Or if you missed Jasper Loy's trick, you can guess and check a value of $a$ for which $$f(a) = a^4 +4a^3 +2a^2 −4a−3 = 0.$$
E.g. f(1) = 0 so $(a-1)$ is a factor and you can use long division to factorise it out.
