College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Review Exercises - Page 75: 71



Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 6(3r-1)^2+(3r-1)-35 ,$ use substitution to make the given expression a trinomial. Then, 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. Finally, use back-substitution to get the factored form of the original expression. $\bf{\text{Solution Details:}}$ Let $z=(3r-1).$ Then, the expression above is equivalent to \begin{array}{l}\require{cancel} 6z^2+z-35 .\end{array} In the trinomial expression above the value of $ac$ is $ 6(-35)=210 $ and the value of $b$ is $ 1 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\{ -14,15 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 6z^2-14z+15z-35 .\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} (6z^2-14z)+(15z-35) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2z(3z-7)+5(3z-7) .\end{array} Factoring the $GCF= (3z-7) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3z-7)(2z+5) .\end{array} Since $z=(3r-1)$, then the expression above is equivalent to \begin{array}{l}\require{cancel} [3(3r-1)-7][2(3r-1)+5] \\\\= [9r-3-7][6r-2+5] \\\\= [9r-10][6r+3] \\\\= [9r-10]3[2r+1] \\\\= 3(9r-10)(2r+1) .\end{array}
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