<s>[INST] Partial fraction decomposition with a nonrepeated irreducible quadratic factor I'm trying to do a partial fraction decomposition on the following rational eqn with a nonrepeated irreducible quadratic factor:
$$\dfrac{-28x^2-92}{(x-4)^2(x^2+1)}$$
I've broken it down into an identiy: $-28x^2 -92 = A((x-4)(x^2-1))+B(x^2+1)+(Cx+D)(x-4)^2$
and solved for B by setting x to 4 ($B-12$), distributed the -12 and moved the resulting quadratic to the left side, leaving me with:
$$12x^2-28x-80 =  A((x-4)(x^2-1)) + (Cx+D)(x-4)^2$$
Now I want to solve for A by setting x to something that will make the coefficient on $Cx+D$ zero, but I'm stumped - setting it to 4 also gets rid of my A. What do I do? Or have I messed up somewhere along the way?
 [/INST] This is what you have done...
$$\dfrac{-28x^2-92}{(x-4)^2(x^2+1)} 
= \dfrac{A}{(x-4)} + \dfrac{B}{(x-4)^2} + \dfrac{Cx+D}{x^2+1}$$
$$-28x^2 -92 = A(x-4)(x^2+1) + B(x^2+1) + (Cx+D)(x-4)^2$$
Letting $x=4$, we get
$$-540 =17B$$
$$B = -\dfrac{540}{17}$$
What I would do now...
Let $B = -\dfrac{540}{17}$ and simplify.
\begin{align}
   17(-28x^2 -92) &= 17A(x-4)(x^2+1) - 540(x^2+1) + 17(Cx+D)(x-4)^2 \\
   64(x^2-16) &= 17A(x-4)(x^2+1) + 17(Cx+D)(x-4)^2 \\
   64(x+4) &= 17A(x^2+1) + 17(Cx+D)(x-4) \\
\end{align}
Again, let $x=4$...
$$512 = 289A$$
$$A = \dfrac{512}{289}$$
Finally. let $A = \dfrac{512}{289}$ and simplify one more time...
\begin{align}
   64(x+4) &= \dfrac{512}{17}(x^2+1) + 17(Cx+D)(x-4) \\
   1088(x+4) &= 512(x^2+1) + 289(Cx+D)(x-4) \\
   -64(8x+15)(x-4) &= 289(Cx+D)(x-4) \\
   -64(8x+15) &= 289(Cx+D) \\
   -\dfrac{512x}{289} - \dfrac{960}{289} &= Cx+D
\end{align}
$$C = -\dfrac{512}{289}$$
$$D = - \dfrac{960}{289}$$
</s>