<s>[INST] 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.
 [/INST] When the denominator contains factors like $(ax^2 + bx + c)^p$ with $b^2 - 4ac < 0$ (such as $x^2 + 2x + 4$ in your case), then the fractions that appear in the expansions are
$$\frac {A_1 x + B_1} {ax^2 + bx + c} + \frac {A_2 x + B_2} {(ax^2 + bx + c)^2} + \dots + \frac {A_p x + B_p} {(ax^2 + bx + c)^p} .$$
In your case, the expansion is
$$ \frac A {x-2} + \frac B {(x-2)^2} + \frac {Cx + D} {x^2 + 2x + 4} + \frac {Ex + F} {(x^2 + 2x + 4)^2} .$$
</s>