Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapters R-5 - Cumulative Review Exercises - Page 364: 27

Answer

$(8w-3z )(2w+7z)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 16w^2+50wz -21z^2 ,$ 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. $\bf{\text{Solution Details:}}$ Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 16(-21)=-336 $ and the value of $b$ is $ 50 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -6,56 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 16w^2-6wz +56wz -21z^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} (16w^2-6wz )+(56wz -21z^2) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2w(8w-3z )+7z(8w -3z) .\end{array} Factoring the $GCF= (8w-3z ) $ of the entire expression above results to \begin{array}{l}\require{cancel} (8w-3z )(2w+7z) .\end{array}
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