Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises - Page 360: 12



Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 6k^2+11k-10 ,$ 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 $ 6(-10)=-60 $ and the value of $b$ is $ 11 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-60\}, \{2,-30\}, \{3,-20\}, \{4,-15\}, \{5,-12\}, \{6,-10\}, \\ \{-1,60\}, \{-2,30\}, \{-3,20\}, \{-4,15\}, \{-5,12\}, \{-6,10\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -4,15 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 6k^2-4k+15k-10 .\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} (6k^2-4k)+(15k-10) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2k(3k-2)+5(3k-2) .\end{array} Factoring the $GCF= (3k-2) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3k-2)(2k+5) .\end{array}
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