Answer
$4(t+1)(t+2)$
Work Step by Step
$\frac{4t+4}{t-3}\times(t^2-t-6)$
$=\frac{4(t+1)}{t-3}\times(t^2+2t-3t-6)$
$=\frac{4(t+1)}{t-3}\times[t(t+2)-3(t+2)]$
$=\frac{4(t+1)}{t-3}\times(t-3)(t+2)$
$=\frac{4(t+1)(t-3)(t+2)}{t-3}$
$=4(t+1)(t+2)$
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: 23
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