Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Review Exercises: 6

Answer

$(z+1) \left( 3z-1 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ (z+1)(z-4)+(z+1)(2z+3) ,$ get the $GCF.$ Then, divide the given expression and the $GCF.$ Express the answer as the product of the $GCF$ and the resulting quotient. $\bf{\text{Solution Details:}}$ Factoring the $GCF= z+1 ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (z+1) \left( \dfrac{(z+1)(z-4)}{z+1}+\dfrac{(z+1)(2z+3)}{z+1} \right) \\\\= (z+1) \left( (z-4)+(2z+3) \right) \\\\= (z+1) \left( z-4+2z+3 \right) \\\\= (z+1) \left( 3z-1 \right) .\end{array}
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