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

Mar 31, 2017

$\frac{{n}^{2} + n - 2}{n + 2} \cdot \frac{4 n}{n - 1} = 4 n$

#### Explanation:

$\frac{{n}^{2} + n - 2}{n + 2} \cdot \frac{4 n}{n - 1}$

$= \frac{{n}^{2} + 2 n - n - 2}{n + 2} \cdot \frac{4 n}{n - 1}$

$= \frac{n \left(n + 2\right) - 1 \left(n + 2\right)}{n + 2} \cdot \frac{4 n}{n - 1}$

$= \frac{\left(n - 1\right) \left(n + 2\right)}{n + 2} \cdot \frac{4 n}{\left(n - 1\right)}$

$= \frac{\cancel{\left(n - 1\right)} \cancel{\left(n + 2\right)}}{\cancel{\left(n + 2\right)}} \cdot \frac{4 n}{\cancel{\left(n - 1\right)}}$

$= 4 n$