# How do you multiply (m+3)/(4m) div (m^2-9)/(32m^2(m-3))?

May 8, 2015

we are given :
$\frac{m + 3}{4 m}$ / (m^2-9)/(32m^2(m-3)

for multiplying we take the reciprocal of the second term:

$= \frac{m + 3}{4 m}$ $\times$ $32 {m}^{2} \times \frac{m - 3}{{m}^{2} - 9}$

we can write
${m}^{2} - 9$ as $\left(m + 3\right) \left(m - 3\right)$

$= \frac{m + 3}{4 m}$ $\times$ 32m^2xx(m-3)/((m+3)(m-3)

$= \frac{\cancel{m + 3}}{4 m}$ $\times$ 32m^2xxcancel(m-3)/(cancel(m+3)cancel(m-3)

$= \frac{32 {m}^{2}}{4 m}$

$= 8 m$