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I was doing an exercise and ran into a problem with their use of factoring. Here is the problem specific to where the issue occurs:

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

If $(x^2+1)^{-1/2}$ was factored out on the top, shouldn't the amount after factoring be just be $1 - x^2$? I am not seeing where the extra $x^2$ is coming from. Can anyone spot the issue I am having?

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2 Answers 2

up vote 5 down vote accepted

When you factor out a $(x^2+1)^{-\frac{1}{2}}$, the remaining quantity is multiplied by $(x^2+1)^{\frac{1}{2}}$. That is, if we have some expression $$(x^2+1)^a \cdot f(x)$$ where $f(x)$ is some polynomial and $a$ is any number, when we factor out a $(x^2+1)^{-\frac{1}{2}}$ we get $$(x^2+1)^{-\frac{1}{2}}\left((x^2+1)^{a+\frac{1}{2}}\cdot f(x)\right).$$

More concretely in the case of your example, because $\frac{1}{2}=-\frac{1}{2}+1$, we have that $$(x^2+1)^{\frac{1}{2}}=(x^2+1)^{-\frac{1}{2}}(x^2+1)^1,$$ and therefore $$\begin{align} (x^2+1)^{\frac{1}{2}}-x^2(x^2+1)^{-\frac{1}{2}} =&(x^2+1)^{-\frac{1}{2}}(x^2+1)^1-x^2(x^2+1)^{-\frac{1}{2}}\\ =& (x^2+1)^{-\frac{1}{2}}\left((x^2+1)-x^2\right)\end{align}$$

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ah, yea, that makes sense. thanks! –  Matt Jun 10 '11 at 18:43

Suppose you have something like $ a - ba^{2}$ and you want to factor it, you do it as follows: Take $a$ outside and we have $$ a -ba^{2} = a \cdot \Bigl[1 - ba\Bigr]$$

This is what is precisely happening in your question. You have taken out $(x^{2}+1)^{-1/2}$ but note that there is an $x^{2}$ quantity, which needs to be seen.

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