Question

How do you find the product of `(n^2+n-2)/(n+2)*(4n)/(n-1)`?

Answer 1

`(n^2+n-2)/(n+2)*(4n)/(n-1)=4n`

Explanation:

`(n^2+n-2)/(n+2)*(4n)/(n-1)`

`=(n^2+2n-n-2)/(n+2)*(4n)/(n-1)`

`=(n(n+2)-1(n+2))/(n+2)*(4n)/(n-1)`

`=((n-1)(n+2))/(n+2)*(4n)/((n-1))`

`=(cancel((n-1))cancel((n+2)))/cancel((n+2))*(4n)/(cancel((n-1)))`

`=4n`