Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.2 - Factoring Trinomials - 5.2 Exercises - Page 337: 17



Work Step by Step

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