How to find the following limit algebraically? I've been trying to answer this for a while and I know it's a simple question relative to most questions that are posted here.
$$
\lim_{x\rightarrow -2}\: \frac{x^4+5x^3+6x^2}{x^2(x+1)-4(x+1)}
$$
If we substitute -2 for $ x $ we get $ 0/0 $, an indeterminate form. I figured that the denominator can be rewritten as $ (x^2-4)(x+1) $. And then I tried to factor something in the numerator but couldn't see anything interesting. How do I find the limit algebraically?
I know that the answer is supposed to be 1, but I don't know how they got there.
Thanks in advance!

General tip (update) : When you can see that the denominator is equal to zero for a value $x=a$ which is the $x\to a$ of the limit, then you should try factoring on both the numerator and the denominator the factor $(x-a)$, such as you can get rid of the $\frac{0}{0}$ issue. This particular example though can also be done by following a row factorization.
Factor the expression at the numerator by taking out $x^2$ and then forming a quadratic factorization inside, as : 
$$x^4+5x^3+6x^2 = x^2(x^2+5x+6) =x^2(x^2+2x+3x+6) = x^2(x+2)(x+3)$$
Then, the given limit is :
$$\lim_{x\rightarrow -2} \frac{x^4+5x^3+6x^2}{x^2(x+1)-4(x+1)} = \lim_{x\to -2} \frac{x^2(x^2 + 5x + 6)}{(x^2-4)(x+1)} = \lim_{x \to -2} \frac{x^2(x+3)(x+2)}{(x-2)(x+2)(x+1)} $$
$$=$$
$$\lim_{x\to -2} \frac{x^2(x+3)}{(x-2)(x+1)} = 1$$
