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.

Often, a problem is handed to us in a slightly convenient form. Here, we may note the quadratic form: 
$$(a^2+2a)^2−2(a^2+2a)−3$$
We can guess it will factor into four factors, so let's find the four roots, via setting the equation to zero and solving. 
$$(a^2+2a)^2−2(a^2+2a)−3=0$$
Lets complete the square: $[(a^2+2a)-1]^2=4$, then $a^2+2a-1=\pm2$. We solve the two equations, $a^2+2a-3=0$ and $a^2+2a+1=0$. This will give real roots, so we can completely factor the above polynomial. I'll let you finish. 
