Partial fraction expansion with quadratic factors in the denominator Question: expand in partial fractions:
$$\frac {x^5+x^4+3x^3-8x^2+28x+48} {x^6-16x^3+64} .$$
I factored the denominator as $(x-2)^2 (x^2+2x+4)^2$.
With a denominator like $(x-1)(x-2)^2$ I know it will be: 
$\frac A {x-1} + \frac B {x-2} + \frac C {(x-2)^2}$ (first of all I don't get why that is?).
But in this exercise, will $\frac {x^5+x^4+3x^3-8x^2+28x+48} {x^6-16x^3+64}$ be equal to $\frac A {x-2} + \frac B {(x-2)^2} + \frac C {x^2+2x+4} + \frac D {(x^2+2x+4)^2}$?
Thanks in advance.

The question is not entirely clear. Your expression can be also writen as
$$
\frac{3 x}{\left(x^2+2 x+4\right)^2}+\frac{x+2}{x^2+2 x+4}+\frac{1}{(x-2)^2}
$$
hope this helps
