# 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!

• Why don't you format your math? Or, perhaps, even talk about how your "normal methods" don't work? – Robin Goodfellow Oct 9 '14 at 9:06
• Welcome to Math SE ! I suppose it looks incredibly hard as long as you did not try ! I suggest you show your efforts. Explain what you tried and tell where you are stuck. – Claude Leibovici Oct 9 '14 at 9:08
• You can plug some values to $x$ into the denominator and see whether it has a root. – John Oct 9 '14 at 9:09
• $x^5+2x^4+x^3-x^2-2x-1$ seems to be zero at $x=\pm1$ so should be relatively easy to factorise. – Henry Oct 9 '14 at 9:11

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.

• (Upvoted!) Nicee Solution. – M.S.E Oct 9 '14 at 9:29

Startup Help

The denominator has a root $x=1$ and $x=-1$ so $$(x-1)(x+1)g(x)=x^5+2x^4+x^3-x^2-2x-1$$

$$(x^2-1)g(x)=x^5+2x^4+x^3-x^2-2x-1$$

Use Long division to find $g(x)$ And factorize further.

• Additional hint: long division should be a last resort here. Look at the symmetry of the coefficients. – David H Oct 9 '14 at 9:17
• @DavidH , Im trying to guess more roots without long division :) – M.S.E Oct 9 '14 at 9:23
• @Tharindu By the Rational Root Theorem, there can be no rational roots of the denominator other than $\pm 1$. – Travis Willse Oct 9 '14 at 9:43
• @Travis, Yep none of my guesses work.... – M.S.E Oct 9 '14 at 9:45
• That's okay, using CiaPan's answer we can factor into a product of quadratic and linear factors very efficiently, the your observation that $\pm 1$ are roots can help with that factorization too. – Travis Willse Oct 9 '14 at 9:47

$$f(x) = \frac{2x^5+15x^4+15x^3+2x^2+2}{x^5+2x^4+x^3-x^2-2x-1}$$

First: By polynomial long division, we obtain $$f(x)= \frac{2x^5+15x^4+15x^3+2x^2+2}{x^5+2x^4+x^3-x^2-2x-1} = 2 + \frac{5x^4 + 13x^3+ 4x^2 + 4x+4}{x^5 + 2x^4 +x^3-x^2-2x - 1}$$

Now, we factor the denominator. Noting that both $x=-1$ and $x = 1$ are roots of the denominator, we know that $$x^5+2x^4 + x^3 - x^2 - 2x - 1 = (x-1)(x+2)g(x)$$ To solve for the third degree polynomial $g(x)$, we divide the denominator by $(x-1)(x+1) = x^2 - 1$ to obtain $$g(x) = x^3 + 2x^2 + 2x +1 = (x+1)(x^2 + x+1)$$

That gives us $$f(x) = 2 +\frac{5x^4+13x^3 + 4x^2 + 4x + 4}{(x-1)(x+1)^2(x^2 + x + 1)} = 2 + \frac{A}{x-1} + \frac{B}{x+1} + \frac C{(x+1)^2} + \frac{Dx + E}{x^2 + x + 1}$$

Can you take it from here? We need only solve for $A, B, C, D, E$, knowing that $$A(x+1)^2(x^2 + x+1) + B(x-1)(x+1)(x^2 + x + 1) + C(x-1)(x^2 + x + 1) + (Dx+E)(x-1)(x+1)^2 = 5x^4 + 13x^3 + 4x^2 + 4x + 4$$