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Find the partial fractions for $\frac{3x^5+7x^4-7x^3-3x^2+x}{3x^2+10x+3}$

First, I recognize it's not a proper fraction. I also know the denominator is $(x+3)(3x+1)$. Now every time I try long division to convert to a proper fraction by hand I get $x^3-x^2+1+(\frac{-10x-3}{(x+3)(3x+1)})$, but this is wrong because I'm supposed to get $x^3-x^2+\frac{x}{(x+3)(3x+1)}$.

I can't find the partial fraction just yet if I can't convert this into a proper fraction. When I tried I got very close, but not exactly correct at all. I wish I could show my work, but it seems displaying long division or synthetic division isn't doable on this website.

Can anyone help? Or provide an alternative method that may not require so much long division?

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Here is the division using polylongdiv in LaTeX.

enter image description here

In case you are interested, here is the LaTeX code for it:

\documentclass{article}
\usepackage{polynom}
\begin{document}

\polylongdiv{3x^5+7x^4-7x^3-3x^2+x}{3x^2+10x+3}

\end{document}  

And at this link you can instead find a WolframAlpha computation of the decomposition.

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  • $\begingroup$ Somehow I kept making the same addition/sign error in the middle of the division. Jesus, now i'm wondering how I didn't see it before. $\endgroup$ – Lex_i Jul 23 at 4:05
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The first step you need to follow is the long division method, which will for sure give you $$\frac{3x^5+7x^4-7x^3-3x^2+x}{3x^2+10x+3}=\frac{(x^3-x^2)(3x^2+10x+3)+x}{(x+3)(3x+1)}=x^3-x^2+\frac{x}{(x+3)(3x+1)}$$Half work done!! Now we just need to break $\frac{x}{(x+3)(3x+1)}$ into partial fractions. That's easy,$$\frac{x}{(x+3)(3x+1)}=\frac{\frac{3}{8}(3x+1)-\frac{1}{8}(x+3)}{(x+3)(3x+1)}=\frac{3}{8(x+3)}-\frac{1}{8(3x+1)}$$Thus, we have$$\frac{3x^5+7x^4-7x^3-3x^2+x}{3x^2+10x+3}=x^3-x^2+\frac{3}{8(x+3)}-\frac{1}{8(3x+1)}$$

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