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 - 5.6 Factoring: A General Strategy - 5.6 Exercise Set - Page 344: 21



Work Step by Step

Factoring the $GCF= 6 $, then the given expression, $ 6y^2+18y-240 $ is equivalent to \begin{array}{l} 6(y^2+3y-40) .\end{array} The two numbers whose product is $ac= 1(-40)=-40 $ and whose sum is $b= 3 $ are $\{ 8,-5 \}$. Using these two numbers to decompose the middle term of the trinomial above, then the complete factored form is \begin{array}{l} 6(y^2+8y-5y-40) \\\\= 6[(y^2+8y)-(5y+40)] \\\\= 6[y(y+8)-5(y+8)] \\\\= 6[(y+8)(y-5)] \\\\= 6(y+8)(y-5) .\end{array}
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