Partial Fractions - $\frac{x^3}{x^2 + 12x +36}$ Ok, so I know that since the numerator has a higher power that long division is needed. So after doing that, the main fraction is $\frac{-6x-36}{x^2 + 12x + 36}$. I think that's right. But my problem is that after you factor the denominator they're both equal to $(x+6)$. So how would you use partial fraction decomposition? 

\begin{array}{rcrrrrrrr}
           &    &  x & -12 \\
           &    & -- & --- &    --- &   --- & \\
x^2+12x+36 & )  & x^3 \\
           &    & x^3 &  12x^2 &   36x \\
           &    &  -- &    --- &   --- \\
           &    &     & -12x^2 &  -36x \\ 
           &    &     & -12x^2 & -144x & -432 \\
           &    &     &    --- &   --- &  --- \\
           &    &     &        &  108x &  432 \\
\end{array}
And so
$$\dfrac{x^3}{(x+6)^2} = x - 12 + \dfrac{108x+432}{(x+6)^2}$$
where 
\begin{align}
   \dfrac{108x+432}{(x+6)^2} &= \dfrac{A}{x+6} + \dfrac{B}{(x+6)^2} \\
   108x + 432 &= A(x+6) + B
\end{align}
$$ \text{Let $x=-6$ and you get $B = -216$.}$$
$$ \text{Let $B = -216$ and you get}$$
\begin{align}
   108x + 432 &= A(x+6) - 216 \\
   A(x+6) &= 108x + 648 \\
   A &= 108
\end{align}
And so
$$\dfrac{x^3}{(x+6)^2} = x - 12 + \dfrac{108}{x+6} - \dfrac{216}{(x+6)^2}$$
