#### Answer

$\frac{2c(c+2)}{(c-1)}$

#### Work Step by Step

To multiply, we need to factor the numerator of the second expression and the denominator of the first.
$\frac{4c}{2c+2}\times\frac{c^2+3c+2}{c-1}$ = $\frac{4c}{(2)(c+1)}\times\frac{(c+2)(c+1)}{c-1}$
Now we can cancel out the $c+1$ factor, as well as 2 from the first numerator and denominator of the first expression, and continue multiplying:
$\frac{4c}{2(c+1)}\times\frac{(c+2)(c+1)}{c-1}$ = $\frac{(2c)}{1}\times\frac{(c+2)}{c-1}$ = $\frac{2c(c+2)}{(c-1)}$