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: 32

Answer

$y(4y-1)(6y+1)$

Work Step by Step

Factoring the $GCF=y$, then the given expression, $ 24y^3-2y^2-y $, is equivalent to \begin{array}{l}\require{cancel} y(24y^2-2y-1) .\end{array} The two numbers whose product is $ac= 24(-1)=-24 $ and whose sum is $b= -2 $ are $\{ -6,4 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ y(24y^2-2y-1) $, is \begin{array}{l}\require{cancel} y(24y^2-6y+4y-1) \\\\= y[(24y^2-6y)+(4y-1)] \\\\= y[6y(4y-1)+(4y-1)] \\\\= y[(4y-1)(6y+1)] \\\\= y(4y-1)(6y+1) .\end{array}
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