calculus - Finding $limlimits_{xrightarrow 2} dfrac{x^3-8}{x^2-x-2}$

I've done so many limit problems in calculus lately, but I can't wrap my mind around how to simplify this one in order to solve it:

$$\lim_{x\rightarrow 2} \dfrac{x^3-8}{x^2-x-2}$$

I understand the $x^3-8$ factors down to $(x-2)(x^2+2x+4)$, but that still leaves us with
$$\lim_{x\rightarrow 2} \dfrac{(x-2)(x^2+2x+4)}{x^2-x-2},$$
which I can't seem to find a way to simplify so that the denominator is not equal to 0.

In case anyone figures out themselves, the answer is 4 (I was given the answer - this is on a review sheet for an upcoming exam). Also, I tagged this as homework, even though it is not technically homework.

So if anyone could help point me in the right direction here, that would be very helpful.

Answer

Hint: Note that $x^2-x-2=(x-2)(x+1)$.