# How do you divide (x^2-9)/(4x+12)div(x-3)/6?

May 24, 2015

$\frac{\left(x - 3\right) \left(x + 3\right)}{2 \left(x + 6\right)} . \frac{6}{x - 3}$ = $\frac{6 \left(x + 3\right)}{2 \left(x + 6\right)}$

May 24, 2015

First, notice that ${x}^{2} - 9 = {x}^{2} - {3}^{2} = \left(x - 3\right) \left(x + 3\right)$
This comes from a standard identity: $\left({a}^{2} - {b}^{2}\right) = \left(a - b\right) \left(a + b\right)$

So $\frac{{x}^{2} - 9}{x - 3} = \frac{\left(x - 3\right) \left(x + 3\right)}{x - 3} = \left(x + 3\right)$

Hence

$\frac{{x}^{2} - 9}{4 x + 12} \div \frac{x - 3}{6}$

$= \frac{x + 3}{4 x + 12} \div \frac{1}{6} = \frac{6 \left(x + 3\right)}{4 \left(x + 3\right)} = \frac{6}{4} = \frac{3}{2}$