Q:

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?

A:

To find out an equation of A and c compare coefficient of $x^3$.put x=0 and compare coefficient of $x^2$.after doing these all operation you will find equation and some values.If you stuck then leave a comment.I'll do it.
 you have this eqn
$$-28x^2-92=A(x-4)(x^2+1)+B(x^2+1)+(Cx+D)(x-4)^2$$
then putting $x=4$ you will get $B=\dfrac{-540}{17}$
compare coefficient of $x^3$ you will get : $0=A+C$
compare coefficient of $x^2$ you will get : $-28=-4A+B-8C+D$
putting  $x=0$ you will get: $-92=-4A+B+16D$
you have value of B so these eqn will become litte more  easy and I hope you will take it to final from now on.

