Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.6 Factoring: A General Strategy - 5.6 Exercise Set: 61

Answer

$4c(5d-c)(4d-c)$

Work Step by Step

Factoring the $GCF= 4c $, then the given expression, $ 80cd^2-36c^2d+4c^3 $ is equivalent to \begin{array}{l} 4c(20d^2-9cd+c^2) .\end{array} The two numbers whose product is $ac= 20(1)=20 $ and whose sum is $b= -9 $ are $\{ -4,-5 \}$. Using these two numbers to decompose the middle term of the expression, $ 4c(20d^2-4cd-5cd+c^2) ,$ then the factored form is \begin{array}{l} 4c[(20d^2-4cd)-(5cd-c^2)] \\\\= 4c[4d(5d-c)-c(5d-c)] \\\\= 4c[(5d-c)(4d-c)] \\\\= 4c(5d-c)(4d-c) .\end{array}
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