# Am I right in thinking $\frac{x^{2}}{ax+b}$ is an improper rational expression?

Am I right in thinking $\dfrac{x^{2}}{ax+b}$ is an improper rational expression? If so, can someone help me figure out how to write it as the sum of a polynomial and proper rational expression?

I have not a clue.

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I'll do the first few steps; here's hoping you'll catch on to what I'm doing:

\begin{align*} \frac{x^2}{ax+b}&=\frac{ax^2}{a(ax+b)}\\ &=\frac{ax^2}{a(ax+b)}+\frac{bx}{a(ax+b)}-\frac{bx}{a(ax+b)}\\ &=\frac{(ax+b)x}{a(ax+b)}-\frac{bx}{a(ax+b)}\\ &=\frac{x}{a}-\frac{b}{a}\frac{x}{ax+b}\\ &=\frac{x}{a}-\frac{b}{a}\left(\frac{ax}{a(ax+b)}+\frac{b}{a(ax+b)}-\frac{b}{a(ax+b)}\right)\\ \end{align*}

Can you take it from here?

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There is more?!?! I stopped at the 4th line! – Magpie Aug 5 '12 at 5:15
Yes.$\phantom{}$ – J. M. Aug 5 '12 at 5:22
I hope my latex is right as I can't see it... Is it $\frac{x}{a}-\frac{b}{a}(\frac{1}{a}-\frac{b}{a(ax+b)})$ – Magpie Aug 5 '12 at 5:32
Good. Multiply it out and you're done. – J. M. Aug 5 '12 at 5:34

Hint: Use polynomial long division to divide $x^2$ by $ax+b$ and you will get a result of the form $(ax+b)(P)+R$. Now divide through by $ax+b$ and you have the desired form.

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When in doubt, try the conjugate. – Alexander Gruber Aug 5 '12 at 4:17
@AlexanderNikolasGruber You should expand on that and make it an answer! :) – Galois Group Aug 5 '12 at 4:19