The only problem with finding the limit- and the only reason why you want to factor- is that both numerator and denominator are 0 when x= -12. And that tells you that (x- (-2))= x+ 2 is a factor. Divide $\displaystyle 4x^3+ 8x^2+ 8x+ 16$ by x+ 2 ("synthetic division" works nicely here) to find the other factor.
note ...
$\displaystyle \frac{4x^3+8x^2+8x+16}{x+2} = \frac{4x^2(x+2) + 8(x+2)}{x+2} = 4x^2+8$
to be continuous at $\displaystyle x = -2$, $\displaystyle \lim_{x \to -2} f(x)$ must exist ...
$\displaystyle \lim_{x \to -2^+} x^2 - 6x + a = \lim_{x \to -2^-} 4x^2 + 8$
$\displaystyle a =$ ?