# How do you simplify -(5c^2d^5)/(8cd^5f^0)?

Mar 5, 2017

See the entire simplification process below:

#### Explanation:

First, use this rule of exponents to eliminate the $f$ term: ${a}^{\textcolor{red}{0}} = 1$

$- \frac{5 {c}^{2} {d}^{5}}{8 c {d}^{5} {f}^{\textcolor{red}{0}}} = - \frac{5 {c}^{2} {d}^{5}}{8 c {d}^{5} 1} = - \frac{5 {c}^{2} {d}^{5}}{8 c {d}^{5}}$

Next, cancel like terms:

$- \frac{5 {c}^{2} {d}^{5}}{8 c {d}^{5}} = - \frac{5 {c}^{2} \textcolor{red}{\cancel{\textcolor{b l a c k}{{d}^{5}}}}}{8 c \textcolor{red}{\cancel{\textcolor{b l a c k}{{d}^{5}}}}} = - \frac{5 {c}^{2}}{8 c}$

Now, use these two rules for exponents to complete the simplification:

$a = {a}^{\textcolor{b l u e}{1}}$ or ${a}^{\textcolor{b l u e}{1}} = a$ and ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$

$- \frac{5 {c}^{2}}{8 c} = - \frac{5 {c}^{\textcolor{red}{2}}}{8 {c}^{\textcolor{b l u e}{1}}} = - \frac{5 {c}^{\textcolor{red}{2} - \textcolor{b l u e}{1}}}{8} = - \frac{5 {c}^{1}}{8} = - \frac{5 c}{8}$