Partial fraction decomposition of a rational function with denominator $x^5+2x^4+x^3-x^2-2x-1$ 
Factorize the denominator completely and write $f(x)$ as a partial fraction given $$f(x) = \frac{2x^5+15x^4+15x^3+2x^2+2}{x^5+2x^4+x^3-x^2-2x-1}$$

Any ideas for this partial fraction question? Not a clue how to go about it, the normal methods Im used to dont work. Any tips for direction will be greatly appreciated
I tried the general methods of breaking down the denominator and placing the parts over A and B, as well as C - the 2 factorizations that I tried using were (x + 1)^2 (x^3 - 1) and (x+1)^2 ( x^2 + x + 1) (x-1). I tried long division of the polynomials as well but for some reason I wasnt able to get it to work. Thanks for the feed back! 

Doesn't a symmetry suggest you something...?
$$x^5+2x^4+x^3-x^2-2x-1$$
$$=(x^5+2x^4+x^3)-(x^2+2x+1)$$
$$=x^3(...?...)-(x^2+2x+1)$$
$$=(x^3-...?...)(x^2+2x+1)$$
Now a symmetry of the last parentheses may be exploited, and $(x^3-\text{const})$ can be further decomposed, too.
EDIT
Before factorisation do the division. Both numerator and denominator have the largest power term $x^5$, so there is an integer part equal $2$ (Why?):
$$\frac{2x^5+15x^4+15x^3+2x^2+2}{x^5+2x^4+x^3-x^2-2x-1} = 2 + \frac{\text {some 4th-degree polynomial}}{x^5+2x^4+x^3-x^2-2x-1}$$
Find the new numerator by subtracting twice the denominator from the original numerator.
Then start breaking your function into partial fractions.
