# Question about factoring polynomial fraction

I ran into this problem in the review section of my math text but I'm not sure how to go about solving it.

$$\frac{2x(x^2-9) - x^2(2x)}{(x^2-9)^2} = 0$$

I can't find a way to cancel the numerator with $(x^2 - 9)$. I guess I could multiply both sides of the equation by $(x^2 - 9)^2$ but I figured that wouldn't be allowed because so much of the expression would be 'lost'. Am I wrong? Could somebody point me in the right direction?

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One should keep track of any restrictions on the value of $x$ due to assumptions such as $(x^2-9)^2 \ne 0$. It turns out not to matter in the end, in this case, but sometimes it does matter. –  David K Jun 4 '14 at 20:10

Nothing will be lost, in fact, you would have then simplified your problem a lot. It is worth noting that if $Q(x)$ and $P(x)$ are two polynomials such that $Q(x)\neq0$ then $$\frac{P(x)}{Q(x)}=0\iff P(x)=0.$$ Therefore your problem becomes: $$\frac{2x(x^2-9) - x^2(2x)}{(x^2-9)^2} = 0\iff 2x(x^2-9) - x^2(2x)=0 \ \ \underline{\underline{\text{and}}} \ \ (x^2-9)^2\neq0.$$ How to move forward? Well, note that you have a common factor which is $2x$, then...
$\frac{2x(x^2-9)-x^2(2x)}{(x^2-9)^2}=0$ if and only if $2x(x^2-9)-x^2(2x)=0$ and $x\ne\pm 3$. Solve.