Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises - Page 348: 68



Work Step by Step

$\bf{\text{Solution Outline:}}$ Group the terms of the given expression, $ 16x^3+32x^2-9x-18 ,$ such that the factored form of the groupings will result to a factor that is common to the entire expression. Then, factor the $GCF$ in each group. and factor the $GCF$ of the entire expression. Finally, factor the resulting difference of $2$ squares. $\bf{\text{Solution Details:}}$ Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (16x^3+32x^2)-(9x+18) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 16x^2(x+2)-9(x+2) .\end{array} Factoring the $GCF= (x+2) $ of the entire expression above results to \begin{array}{l}\require{cancel} (x+2)(16x^2-9) .\end{array} The expressions $ 16x^2 $ and $ 9 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 16x^2-9 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x+2)[(4x)^2-(3)^2] \\\\= (x+2)[(4x+3)(4x-3)] \\\\= (x+2)(4x+3)(4x-3) .\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.