# How do you multiply and simplify \frac { 3p ^ { 4} } { 5q ^ { 5} ( r - 5) ^ { 3} } \cdot \frac { 4l q ^ { 2} ( r - 5) } { 21p ^ { 3} }?

Apr 17, 2017

$\frac{4 p l}{35 {q}^{3} {\left(r - 5\right)}^{2}}$

#### Explanation:

$\frac{\textcolor{g r e e n}{3} {p}^{\textcolor{red}{4}}}{5 {q}^{5} {\left(r - 5\right)}^{3}} \cdot \frac{4 l {q}^{2} \left(r - 5\right)}{\textcolor{g r e e n}{21} {p}^{\textcolor{red}{3}}}$

Subtract the smaller exponent of a base from both bases, because of the division, and simplify $3$ and $21$ by dividing them by $3$

$\frac{\cancel{3} {p}^{4 - 3}}{5 {q}^{5} {\left(r - 5\right)}^{3}} \cdot \frac{4 l {q}^{2} \left(r - 5\right)}{7 \cancel{21} \cancel{{p}^{3 - 3}}}$

Anything to the power of zero can be cancelled

$\left(r - 5\right)$ is the same as ${\left(r - 5\right)}^{1}$

$= \frac{p}{5 {q}^{5} {\left(r - 5\right)}^{\textcolor{red}{3}}} \cdot \frac{4 l {q}^{2} {\left(r - 5\right)}^{\textcolor{red}{1}}}{7}$

$= \frac{p}{5 {q}^{5} {\left(r - 5\right)}^{\textcolor{red}{3} - 1}} \cdot \frac{4 l {q}^{2} \cancel{{\left(r - 5\right)}^{\textcolor{red}{1} - 1}}}{7}$

$= \frac{p}{5 {q}^{\textcolor{red}{5}} {\left(r - 5\right)}^{2}} \cdot \frac{4 l {q}^{\textcolor{red}{2}}}{7}$

$= \frac{p}{5 {q}^{\textcolor{red}{5} - 2} {\left(r - 5\right)}^{2}} \cdot \frac{4 l \cancel{{q}^{\textcolor{red}{2} - 2}}}{7}$

$= \frac{p}{5 {q}^{3} {\left(r - 5\right)}^{2}} \cdot \frac{4 l}{7}$

Now there is no other simplification, we will multiply these two expressions together

$= \frac{p \cdot 4 l}{\textcolor{g r e e n}{5} {q}^{3} {\left(r - 5\right)}^{2} \cdot \textcolor{g r e e n}{7}}$

$= \frac{4 p l}{35 {q}^{3} {\left(r - 5\right)}^{2}}$

After a final check, this is the most simplified form.