Answer
$=\frac{2c^2+4c}{c-1}$
with $c \ne 1 $
Work Step by Step
$\frac{4c}{2c+2}.\frac{c^2+3c+2}{c-1}$
Multiply
$=\frac{4c}{2(c+1)}.\frac{(c+1)(c+2)}{c-1}$
$=\frac{2c(c+2)}{c-1}$
$=\frac{2c^2+4c}{c-1}$
with $c \ne 1 $
Algebra 1: Common Core (15th Edition)
by Charles, Randall I.
Published by
Prentice Hall
ISBN 10:
0133281140
ISBN 13:
978-0-13328-114-9
Chapter 11 - Rational Expressions and Functions - 11-2 Multiplying and Dividing Rational Expressions - Practice and Problem-Solving Exercises - Page 674: 17
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