Limit on a five term polynomial it has been two years since I have taken or used calculus, and I am having some trouble with factoring a polynomial in order to take a limit on it. I have searched for previous similar questions here, but I have been unable to find anything helpful. Here is my problem:
I have:
\begin{equation} 
\lim_{x \to\ 1}\frac{{x^4+3x^3-13x^2-27x+36}}{x^2+3x-4}
\end{equation}
So, how do I approach this? When I try long division, I end up with a remainder, e.g.:
\begin{equation}
_{x \to\ 4}\frac{{x^4+3x^3-13x^2-27x+36}}{x^2+3x-4} = (x+3)(x+3)+\frac{{24x+72}}{(x+1)(x-4)}
\end{equation}
Plugging in the limit to what I came up with via long division just yields an undefined result, so I'm obviously doing something wrong. 
If anyone can help, I would really appreciate it!
Thanks!

Ok, so I figured out my error. I was allowing the term \begin{equation}13x^2\end{equation} from the numerator to mess me up. Factoring the numerator needs to be approached by first turning the prime number, 13, into non-prime numbers. Thus,\begin{equation}x^4+3x^3-13x^2-27x+36\end{equation}becomes \begin{equation}x^4+3x^3-9x^2-4x^2-27x+36\end{equation}Next, I group the terms that have common factors. Thus the previous term becomes \begin{equation}x^2(x^2+3x-4)-9(x^2+3x-4)\end{equation}Since the term \begin{equation} (x^2+3x-4)\end{equation} is identical to the denominator, the only terms left are \begin{equation}x^2-9\end{equation}Hence, \begin{equation}\lim_{x \to\ 1}x^2-9=-8\end{equation} 
