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: 38

Answer

$(7c+2d)(2c-3d)$

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, $ 14c^2-17cd-6d^2 ,$ the value of $ac$ is $ 14(-6)=-84 $ and the value of $b$ is $ -17 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-84\}, \{2,-42\}, \{3,-28\}, \{4,-21\}, \{7,-12\}, \{-1,84\}, \{-2,42\}, \{-3,28\}, \{-4,21\}, \{-7,12\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 4,-21 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} 14c^2+4cd-21cd-6d^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} (14c^2+4cd)-(21cd+6d^2) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2c(7c+2d)-3d(7c+2d) .\end{array} Factoring the $GCF= (7c+2d) $ of the entire expression above results to \begin{array}{l}\require{cancel} (7c+2d)(2c-3d) .\end{array}
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