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: 32



Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $ (3x+2)(x-4)-(3x+2)(x+8) .$ 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 $\{ (3x+2),(3x+2) \}$ is $ (3x+2) .$ Hence, the entire expression has $GCF= (3x+2) .$ Factoring the $GCF= (3x+2) ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} (3x+2) \left( \dfrac{(3x+2)(x-4)}{(3x+2)}-\dfrac{(3x+2)(x+8)}{(3x+2)} \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} (3x+2) \left( (3x+2)^{1-1}(x-4)-(3x+2)^{1-1}(x+8) \right) \\\\= (3x+2) \left( (3x+2)^{0}(x-4)-(3x+2)^{0}(x+8) \right) \\\\= (3x+2) \left( (1)(x-4)-(1)(x+8) \right) \\\\= (3x+2) \left( x-4-x-8 \right) \\\\= (3x+2) \left( -12 \right) \\\\= -12(3x+2) .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.