# How do you divide  {(12m^2n^5)/(m+5)}/{(3m^3n)/(m^2-25)}?

Mar 6, 2016

The answer is $4 {m}^{-} 1 {n}^{4} \left(m - 5\right)$ Or $\frac{4 {n}^{4} \left(m - 5\right)}{m}$

#### Explanation:

Basically what you are dealing with here is a fraction,
$\frac{12 {m}^{2} {n}^{5}}{m + 5}$ , being divided by a second fraction

$\frac{3 {m}^{3} n}{{m}^{2} - 25}$. So how do we divide fractions? One simple

technique we can use is to multiply the first fraction by the

reciprocal of the second fraction. So our solution looks like this:

$\left(\frac{12 {m}^{2} {n}^{5}}{m + 5}\right)$ $\left(\frac{{m}^{2} - 25}{3 {m}^{3} n}\right)$ = (12m^2n^5(m^2-25))/(3m^3n(m +5).
While this is the solution, it is not yet reduced to simplest terms, which is our next step. Notice that in the numerator, $\left({m}^{2} - 25\right)$ is a perfect square binomial, which factors easily into $\left(m + 5\right) \left(m - 5\right)$. So our original solution from above can now be factored into:
$\frac{\left(3\right) \left(4\right) \left({m}^{2}\right) \left(n\right) \left({n}^{4}\right) \left(m + 5\right) \left(m - 5\right)}{3 \left({m}^{2}\right) \left(m\right) \left(n\right) \left(m + 5\right)}$. We can now divide or "cancel" like terms to arrive at the simplified answer of $\frac{4 {n}^{4} \left(m - 5\right)}{m}$.