How do I solve this Partial Fractions question? $$\frac{x^4}{(x^2-3)(x^2+3)}\;$$  
How would I do this? 
My attempt started with this:  
$$\frac{A+Bx}{(x^2-3)}\; + \frac{C+Dx}{(x^2+3)}$$  
But when I start working it all out, A, B, C and D all go to 0. What am I doing wrong? 

Now that the $x$ in the numerator is really supposed to be $x^4$, we first need the degree in the numerator to be less than that of the denominator before employing partial fraction decomposition. Note that the denominator can be expressed as the difference of squares: $x^4 - 9$, which makes polynomial division very straightforward. We can also see the factor $x^2 - 3$ as a difference of squares and factor accordingly knowing $x^2 - 3 = (x + \sqrt 3)(x - \sqrt 3)$:
$$\begin{align} \frac{x^4}{(x^2-3)(x^2+3)} & = \dfrac{(x^4 - 9) + 9}{(x^2 - 3)(x^2 + 3)} \\ \\\dfrac {x^4 - 9}{x^4 - 9} + \dfrac 9{(x^2 - 3)(x^2 + 3)} &= 1 + \dfrac 9{(x+ \sqrt 3)(x - \sqrt 3)(x^2 + 3)} \\ \\ &= 1 + \dfrac A{x + \sqrt 3} + \frac B{x - \sqrt 3} + \frac{Cx + D}{x^2 + 3}\end{align}$$
