Question

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

Answer 1

`= - 6/ (k + 2)`

Explanation:

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

change `(k-1) to -(-k + 1)` then rearrange

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

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

`= 6/- 1 -: (k + 2)/1`

change sign `-: to *` then rearrange the operation of fraction.

`= 6/- 1 * 1/ (k + 2)`

`= - 6/ (k + 2)`

Answer 2

The final expression is `(-6)/(k+2)`.

Explanation:

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

`color(white)=(6(1-k))/(k-1)div(4k^2(k+2))/(4k^2)`

`=(6(-k+1))/(k-1)div(4k^2(k+2))/(4k^2)`

`=(-6(k-1))/(k-1)div(4k^2(k+2))/(4k^2)`

`=(-6color(red)cancelcolor(black)((k-1)))/color(red)cancelcolor(black)(k-1)div(color(red)cancelcolor(black)(4k^2)(k+2))/color(red)cancelcolor(black)(4k^2)`

`=-6div(k+2)`

`=(-6)/(k+2)`

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