Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 6 - Factoring, Solving Equations, and Problem Solving - 6.4 - Factoring Trinomials of the Form ax^2+bx+c - Problem Set 6.4: 10

Answer

$2(3y+4)(y-2)$

Work Step by Step

Since $2$ is common to all the terms of the equation, we take it out as a common factor: $6y^{2}-4y-16$ =$2(3y^{2}-2y-8)$ In accordance with the rules of factoring trinomials, we need to find two integers whose product is $-24$ and whose sum is $-2$. A little searching determines that these numbers are $4$ and $-6$. Therefore, we can express the middle term $-2y$ as $(+4y-6y)$ and proceed to factoring by grouping: $2(3y^{2}-2y-8)$ =$2(3y^{2}+4y-6y-8)$ =$2[y(3y+4)-2(3y+4)]$ =$2[(3y+4)(y-2)]$ =$2(3y+4)(y-2)$
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