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: 37

Answer

$(3a-2b)(5a-4b)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the quadratic expression $ax^2+bx+c,$ find two numbers whose product is $ac$ and whose sum is $b$. Use these $2$ numbers to decompose the middle term of the quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ In the given expression, $ 15a^2-22ab+8b^2 ,$ the value of $ac$ is $ 15(8)=120 $ and the value of $b$ is $ -22 .$ The $2$ numbers that have a product $ac$ and a sum of $b$ are $\{ -10,-12 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} 15a^2-10ab-12ab+8b^2 .\end{array} Grouping the first and third terms and the second and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (15a^2-10ab)-(12ab-8b^2) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5a(3a-2b)-4b(3a-2b) .\end{array} Factoring the $GCF= (3a-2b) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3a-2b)(5a-4b) .\end{array}
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