# How do you divide \frac { 6( 1- k ) } { k - 1} \div \frac { 4k ^ { 2} ( k + 2) } { 4k ^ { 2} )?

Mar 14, 2018

$= - \frac{6}{k + 2}$

#### Explanation:

$\frac{6 \left(1 - k\right)}{k - 1} \div \frac{4 {k}^{2} \left(k + 2\right)}{4 {k}^{2}}$

change $\left(k - 1\right) \to - \left(- k + 1\right)$ then rearrange

$= \frac{6 \left(1 - k\right)}{-} \left(- k + 1\right) \div \frac{4 {k}^{2} \left(k + 2\right)}{4 {k}^{2}}$

= (6(cancel(1 - k)))/- (cancel(1 - k )) -: (cancel(4 k^2)(k + 2))/(cancel (4k^2)

$= \frac{6}{-} 1 \div \frac{k + 2}{1}$

change sign $\div \to \cdot$ then rearrange the operation of fraction.

$= \frac{6}{-} 1 \cdot \frac{1}{k + 2}$

$= - \frac{6}{k + 2}$

Mar 14, 2018

The final expression is $\frac{- 6}{k + 2}$.

#### Explanation:

Factor the expressions, cancel like terms, then write the final division. Here's what it looks like:

$\textcolor{w h i t e}{=} \frac{6 \left(1 - k\right)}{k - 1} \div \frac{4 {k}^{2} \left(k + 2\right)}{4 {k}^{2}}$

$= \frac{6 \left(- k + 1\right)}{k - 1} \div \frac{4 {k}^{2} \left(k + 2\right)}{4 {k}^{2}}$

$= \frac{- 6 \left(k - 1\right)}{k - 1} \div \frac{4 {k}^{2} \left(k + 2\right)}{4 {k}^{2}}$

$= \frac{- 6 \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(k - 1\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{k - 1}}}} \div \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4 {k}^{2}}}} \left(k + 2\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4 {k}^{2}}}}}$

$= - 6 \div \left(k + 2\right)$

$= \frac{- 6}{k + 2}$

This is as simplified as it gets. Hope this helped!