Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set: 65

Answer

$2(2y+9)(8y-3)$

Work Step by Step

Factoring the $GCF= 2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 132y+32y^2-54 \\\\= 32y^2+132y-54 \\\\= 2(16y^2+66y-27) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 2(16y^2+66y-27) \end{array} has $ac= 16(-27)=-432 $ and $b= 66 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 72,-6 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2(16y^2+72y-6y-27) .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} 2[(16y^2+72y)-(6y+27)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2[8y(2y+9)-3(2y+9)] .\end{array} Factoring the $GCF= (2y+9) $ of the entire expression above results to \begin{array}{l}\require{cancel} 2[(2y+9)(8y-3)] \\\\= 2(2y+9)(8y-3) .\end{array}
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