How to show that $\frac{x^2}{x-1}$ simplifies to $x + \frac{1}{x-1} +1$

How does $\frac{x^2}{(x-1)}$ simplify to $x + \frac{1}{x-1} +1$?

The second expression would be much easier to work with, but I cant figure out how to get there.

Thanks

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$$\frac{x^2}{x-1}=\frac{x^2-1+1}{x-1}=\frac{(x-1)(x+1)+1}{x-1}=x+1+\frac1{x-1}$$ – lab bhattacharjee Jan 14 '13 at 10:26
@Alexander I can see this is at least your second question on this site. When you get a satisfatory answer to your questions, you should accept it as an answer. – Git Gud Jan 14 '13 at 10:59
This sort of problem has a universal solution called "the division algorithm" and "Bézout's identity ". – Luqing Ye Jan 14 '13 at 12:12

A general $^1$ method is to perform the polynomial long division algorithm in the following form or another equivalent one:

to get

$$x^{2}=(x-1)(x+1)+1.$$

Divide both sides by $x-1$:

$$\begin{eqnarray*} \frac{x^{2}}{x-1} &=&\frac{(x-1)(x+1)+1}{x-1}=\frac{(x-1)(x+1)}{x-1}+\frac{1}{x-1} \\ &=&\frac{x+1}{1}+\frac{1}{x-1}=x+1+\frac{1}{x-1}=x+\frac{1}{x-1}+1 \end{eqnarray*}$$

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$^1$ The very same method can be applied to find (or show) that

$$\frac{x^{4}-2x^{3}-6x^{2}+12x+15}{x^{3}+x^{2}-4x-4}=x-3+\frac{ x^{2}+4x+3}{x^{3}+x^{2}-4x-4};$$

In this case the polynomial long division algorithm corresponds to the following computation (or another equivalent one):

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Are the signs right in your long division? I've never seen it done like that before. – Peter Phipps Jan 14 '13 at 11:34
@PeterPhipps The shown signs incorporate the minus sign of a difference. I posted this version because it is the output of the \polylongdiv{x^2}{x-1} LaTeX command (\usepackage{polynom}). – Américo Tavares Jan 14 '13 at 11:53
Thanks for that. I've not seen the Polynom package. – Peter Phipps Jan 14 '13 at 12:15

Very clever trick: If you have to show that two expressions are equivalent, you work backwards. \begin{align}=& x +\frac{1}{x-1} + 1 \\ \\ \\ =& \frac{x^2 - x}{x-1} +\frac{1}{x - 1} + \frac{x-1}{x-1} \\ \\ \\ =& \frac{x^2 - x + 1 + x - 1}{x - 1} \\ \\ \\ =&\frac{x^2 }{x - 1}\end{align}Now, write the steps backwards (if you're going to your grandmommy's place, you turn backwards and then you again turn backwards, you're on the right way!) and act like a know-it-all.

\begin{align}=&\frac{x^2 }{x - 1} \\ \\ \\ =& \frac{x^2 - x}{x-1} +\frac{1}{x - 1} + \frac{x-1}{x-1} \\ \\ \\ =& x +\frac{1}{x-1} + 1 \end{align} Q.E.Doodly dee! This trick works and you can impress your friends with such elegant proofs produced by this trick.

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Instructive, but I $+1$ed it for the humor. – Git Gud Jan 14 '13 at 10:45
+1 to emphasize the point that "show this is true" is a different problem than "figure out how to simplify $x^2/(x-1)$". Once you're in the "real world" rather than canned textbook problems, it is not uncommon that you can guess the answer through alternative means and then need to justify it. Eventually, you start solving some problems not by attacking it directly, but by trying to figure out ways to guess the answer! – Hurkyl Jan 14 '13 at 11:32
+1 for the Q.E.Doodly dee. – dwarandae Jan 14 '13 at 16:59

$\textbf{Hint:}$ Do you know about polynomial long divison?

A simpler way of dealing with this problem is noticing that for all $x\neq 1$,

$$\frac{x^2}{x-1}=\frac{x^2-1+1}{x-1}=\frac{x^2-1}{x-1}+\frac{1}{x-1}=\frac{(x-1)(x+1)}{x-1}+\frac{1}{x-1}=x+1+\frac{1}{x-1}.$$

It's not always this easy, though. So you should learn polynomial long divison.

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We have that $$x^2=(x-1+1)^2=(x-1)^2+1+2(x-1)$$ and so for $x\neq 1$: $$\frac{x^2}{x-1}=\frac{(x-1)^2+1+2(x-1)}{x-1}=x-1+\frac{1}{x-1}+2=x+1+\frac{1}{x-1}.$$

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$$\frac{x^2}{x-1}=\frac{x^2-x+x-1+1}{x-1} = \frac{x(x-1) + (x-1)+1}{x-1}=\frac{x(x-1)}{x-1}+\frac{x-1}{x-1}+\frac{1}{x-1}$$

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Hint: An alternative method would be to compute: $$\frac{x^2}{x-1}-\left(x+1+\frac1{x-1}\right)$$ And then conclude that since $$\frac{x^2}{x-1}-\left(x+1+\frac1{x-1}\right)=0$$ Then $$\frac{x^2}{x-1}=x+1+\frac1{x-1}.$$

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