Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises - Page 360: 13



Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 12r^2-5r-3 ,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ To factor the trinomial expression above, note that the value of $ac$ is $ 12(-3)=-36 $ and the value of $b$ is $ -5 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-36\}, \{2,-18\}, \{3,-12\}, \{4,-9\}, \{6,-6\}, \\ \{-1,36\}, \{-2,18\}, \{-3,12\}, \{-4,9\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 4,-9 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 12r^2+4r-9r-3 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (12r^2+4r)-(9r+3) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 4r(3r+1)-3(3r+1) .\end{array} Factoring the $GCF= (3r+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3r+1)(4r-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.