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: 29

Answer

$(m-4) \left( 2m+5 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $ (m-4)(m+2)+(m-4)(m+3) .$ 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 $\{ (m-4),(m-4) \}$ is $ (m-4) .$ Hence, the entire expression has $GCF= (m-4) .$ Factoring the $GCF= (m-4) ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} (m-4) \left( \dfrac{(m-4)(m+2)}{(m-4)}+\dfrac{(m-4)(m+3)}{(m-4)} \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} (m-4) \left( (m-4)^{1-1}(m+2)+(m-4)^{1-1}(m+3) \right) \\\\= (m-4) \left( (m-4)^{0}(m+2)+(m-4)^{0}(m+3) \right) \\\\= (m-4) \left( (1)(m+2)+(1)(m+3) \right) \\\\= (m-4) \left( m+2+m+3 \right) \\\\= (m-4) \left( 2m+5 \right) .\end{array}
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