# How do you simplify (-12c^3d^0f^-2)/(6c^5d^-3f^4)?

Jun 18, 2017

See a solution process below:

#### Explanation:

First, rewrite this expression as:

$\left(- \frac{12}{6}\right) \left({c}^{3} / {c}^{5}\right) \left({d}^{0} / {d}^{-} 3\right) \left({f}^{-} \frac{2}{f} ^ 4\right) \implies$

$- 2 \left({c}^{3} / {c}^{5}\right) \left({d}^{0} / {d}^{-} 3\right) \left({f}^{-} \frac{2}{f} ^ 4\right)$

Next, use this rule of exponents to simplify the $d$ term:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$

$- 2 \left({c}^{3} / {c}^{5}\right) \left({d}^{\textcolor{red}{0}} / {d}^{\textcolor{b l u e}{- 3}}\right) \left({f}^{-} \frac{2}{f} ^ 4\right) \implies$

$- 2 \left({c}^{3} / {c}^{5}\right) \left({d}^{\textcolor{red}{0} - \textcolor{b l u e}{- 3}}\right) \left({f}^{-} \frac{2}{f} ^ 4\right) \implies$

$- 2 \left({c}^{3} / {c}^{5}\right) \left({d}^{\textcolor{red}{0} + \textcolor{b l u e}{3}}\right) \left({f}^{-} \frac{2}{f} ^ 4\right) \implies$

$- 2 \left({c}^{3} / {c}^{5}\right) {d}^{3} \left({f}^{-} \frac{2}{f} ^ 4\right)$

Now, use this rule of exponents to simplify the $c$ and $f$ terms:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = \frac{1}{x} ^ \left(\textcolor{b l u e}{b} - \textcolor{red}{a}\right)$

$- 2 \left({c}^{\textcolor{red}{3}} / {c}^{\textcolor{b l u e}{5}}\right) {d}^{3} \left({f}^{\textcolor{red}{- 2}} / {f}^{\textcolor{b l u e}{4}}\right) \implies$

$- 2 \left(\frac{1}{c} ^ \left(\textcolor{b l u e}{5} - \textcolor{red}{3}\right)\right) {d}^{3} \left(\frac{1}{f} ^ \left(\textcolor{b l u e}{4} - \textcolor{red}{- 2}\right)\right) \implies$

$- 2 \left(\frac{1}{c} ^ 2\right) {d}^{3} \left(\frac{1}{f} ^ \left(\textcolor{b l u e}{4} + \textcolor{red}{2}\right)\right) \implies$

$- 2 \left(\frac{1}{c} ^ 2\right) {d}^{3} \left(\frac{1}{f} ^ 6\right) \implies$

$\frac{- 2 {d}^{3}}{{c}^{2} {f}^{6}}$