Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.1 - Greatest Common Factors and Factoring by Grouping - 5.1 Exercises - Page 329: 30


$(z-5) \left( 2z+17 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $ (z-5)(z+7)+(z-5)(z+10) .$ Divide the given expression and the $GCF.$ Express the answer as the product of the $GCF$ and the resulting quotient. $\bf{\text{Solution Details:}}$ The $GCF$ of the constants of the terms $\{ 1,1 \}$ is $ 1 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variables $\{ (z-5)),(z-5) \}$ is $ (z-5) .$ Hence, the entire expression has $GCF= (z-5) .$ Factoring the $GCF= (z-5) ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} (z-5) \left( \dfrac{(z-5)(z+7)}{(z-5)}+\dfrac{(z-5)(z+10)}{(z-5)} \right) .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} (z-5) \left( (z-5)^{1-1}(z+7)+(z-5)^{1-1}(z+10) \right) \\\\= (z-5) \left( (z-5)^{0}(z+7)+(z-5)^{0}(z+10) \right) \\\\= (z-5) \left( (1)(z+7)+(1)(z+10) \right) \\\\= (z-5) \left( z+7+z+10 \right) \\\\= (z-5) \left( 2z+17 \right) .\end{array}
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