# How do you simplify (x^2-4)/(x-3) * (x+2)/(8x-16)?

Oct 24, 2015

$\frac{{x}^{2} + 4 x + 4}{8 x - 24}$

#### Explanation:

We should start by factoring the numerators and denominators separately and then seeing what we might be able to cancel out. The four terms we have are;

${x}^{2} - 4$
$x - 3$
$x + 2$
$8 x - 16$

$x - 3$ and $x + 2$ can't really be factored any further, but the other two can. Let's start with ${x}^{2} - 4$. For a binomial expression of the form ${x}^{2} - {a}^{2}$;

${x}^{2} - {a}^{2} = \left(x - a\right) \left(x + a\right)$

In our case, $a = 2$, so;

${x}^{2} - 4 = {x}^{2} - {2}^{2} = \left(x - 2\right) \left(x + 2\right)$

Now take a look at $8 x - 16$. We can factor out the $8$, leaving;

$8 \left(x - 2\right)$

We can now rewrite the original expression as;

$\frac{\left(x - 2\right) \left(x + 2\right) \left(x + 2\right)}{\left(x - 3\right) 8 \left(x - 2\right)}$

The only term that cancels out is $x - 2$.

$\frac{\cancel{x - 2} \left(x + 2\right) \left(x + 2\right)}{\left(x - 3\right) 8 \cancel{x - 2}}$

When we multiply the remaining factors together, we get;

$\frac{{x}^{2} + 4 x + 4}{8 x - 24}$