Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.6 - Factoring Trinomials - Exercise Set: 84

Answer

$4x(3y-2z^2)^2$

Work Step by Step

Factoring the $GCF=4x$, then the given expression, $ 36xy^2-48xyz^2+16xz^4 $, is equivalent to \begin{array}{l} 4x(9y^2-12yz^2+4z^4) .\end{array} The two numbers whose product is $ac= 9(4)=36 $ and whose sum is $b= -12 $ are $\{ -6,-6 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ 4x(9y^2-12yz^2+4z^4) $, is \begin{array}{l}\require{cancel} 4x(9y^2-6yz^2-6yz^2+4z^4) \\\\= 4x[(9y^2-6yz^2)-(6yz^2-4z^4)] \\\\= 4x[3y(3y-2z^2)-2z^2(3y-2z^2)] \\\\= 4x[(3y-2z^2)(3y-2z^2)] \\\\= 4x(3y-2z^2)^2 .\end{array}
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