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

1 Answer
Mar 31, 2017

(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