# How do you divide (m^3n^2)/(m^-1n^3)?

Jul 14, 2015

$\frac{{m}^{3} {n}^{2}}{{m}^{- 1} {n}^{3}} = {m}^{4} {n}^{-} 1$

#### Explanation:

First, let's use the distributive rule to separate the monomials.

$\frac{{m}^{3} {n}^{2}}{{m}^{- 1} {n}^{3}} = \left({m}^{3} / {m}^{-} 1\right) \left({n}^{2} / {n}^{3}\right)$

Now we use the quotient rule: when dividing monomials that have the same base, we subtract the exponents.

So

$\left({m}^{3} / {m}^{-} 1\right) \left({n}^{2} / {n}^{3}\right) = \left({m}^{3 - \left(- 1\right)}\right) \left({n}^{2 - 3}\right) = {m}^{3 + 1} {n}^{-} 1$

$\frac{{m}^{3} {n}^{2}}{{m}^{- 1} {n}^{3}} = {m}^{4} {n}^{-} 1$