# How do you multiply ((x+1)^2) / (x-1) * [(2x-2)/ (x+1)]?

Apr 17, 2018

$\frac{{\left(x + 1\right)}^{2}}{x - 1} \cdot \left(\frac{2 x - 2}{x + 1}\right) = 2 x + 2$

#### Explanation:

Simplify FIRST! This makes you life much easier! Notice that the $x + 1$ factor in the denominator of the second quotient CANCELS OUT one of the $x + 1$ factors in the numerator of the first quotient.

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

Now note that we can factor a 2 out of the second factor in the numerator.

$= \frac{\left(2 x - 2\right) \left(x + 1\right)}{x - 1} = \frac{2 \left(x - 1\right) \left(x + 1\right)}{x - 1}$

Now cancel the $x - 1$ factor.

$\frac{2 \cancel{\left(x - 1\right)} \left(x + 1\right)}{\cancel{x - 1}} = 2 \left(x + 1\right) = 2 x + 2$

Note that for this example, there was very little multiplication required to multiply these quotients. Mostly we divided.