## College Algebra (6th Edition)

$\frac{(x+3)(x-3)}{x(x+4)}$; $x\ne-4, 0, 3$
Before we multiply two fractions, we should factor each to see if we can cancel any terms. Top-left: $x^{2}-9$ is a difference of squares and factors to $(x + 3)(x - 3)$. Top-right: We can pull out a greatest common factor of $3$. Bottom-right: This factors as $(x + 4)(x - 3)$. $\frac{(x+3)(x-3)}{x^{2}}\times \frac{x(x-3)}{(x+4)(x-3)}$ Note that we can cancel the $x-3$ terms, as well as an $x$; this eliminates the $x$ from top-right, though we still have a single $x$ in the bottom-left. Nothing else cancels, so we multiply across, leaving us with our answer of $\frac{(x+3)(x-3)}{x(x+4)}$. Also note that this is true only if $x\ne-4, 0, 3$, since they make the original denominators equal to 0.