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 - 5.2 Factoring Trinomials of the Type x2+bx+c - 5.2 Exercise Set - Page 317: 26

Answer

$3(y-7)(y+4)$

Work Step by Step

Factoring the $GCF= 3 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 3y^2-9y-84 \\\\= 3(y^2-3y-28) .\end{array} Using the factoring of trinomials in the form $x^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 3(y^2-3y-28) \end{array} has $c= -28 $ and $b= -3 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -7,4 \right\}.$ Using these two numbers, the $\text{ expression }$ above is equivalent to \begin{array}{l}\require{cancel} 3(y-7)(y+4) .\end{array}
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