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: 25

Answer

$y^2(y-2)(3y+5)$

Work Step by Step

Factoring the $GCF=y^2$ of the given expression, $ 3y^4-y^3-10y^2 $, results to \begin{array}{l}\require{cancel} y^2(3y^2-y-10) .\end{array} The two numbers whose product is $ac= 3(-10)=-30 $ and whose sum is $b= -1 $ are $\{ -6,5 \}$. Using these two numbers to decompose the middle term, then the factored form of the resulting expression, $ y^2(3y^2-y-10) $,is \begin{array}{l}\require{cancel} y^2(3y^2-6y+5y-10) \\\\= y^2[(3y^2-6y)+(5y-10)] \\\\= y^2[3y(y-2)+5(y-2)] \\\\= y^2[(y-2)(3y+5)] \\\\= y^2(y-2)(3y+5) .\end{array}
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