Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set: 10

Answer

$y^3(2x-1)^2$

Work Step by Step

Factoring the $GCF=y^3$, then the given expression, $ 4x^2y^3-4xy^3+y^3 $, is equivalent to \begin{array}{l} y^3(4x^2-4x+1) .\end{array} The two numbers whose product is $ac= 4(1)=4 $ and whose sum is $b= -4 $ are $\{ -2,-2 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ y^3(4x^2-4x+1) $, is \begin{array}{l}\require{cancel} y^3(4x^2-2x-2x+1) \\\\= y^3[(4x^2-2x)-(2x-1)] \\\\= y^3[2x(2x-1)-(2x-1)] \\\\= y^3[(2x-1)(2x-1)] \\\\= y^3(2x-1)^2 .\end{array}
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