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: 68

Answer

$(2y+3)(6y-5)$

Work Step by Step

The given expression, $ 8y-15+12y^2 ,$ can be re-written as \begin{array}{l} 12y^2+8y-15 .\end{array} The two numbers whose product is $ac= 12(-15)=-180 $ and whose sum is $b= 8 $ are $\{ 18,-10 \}$. Using these two numbers to decompose the middle term of the expression, $ 8t^2+14t-15 ,$ then the factored form is \begin{array}{l} 12y^2+18y-10y-15 \\\\= (12y^2+18y)-(10y+15) \\\\= 6y(2y+3)-5(2y+3) \\\\= (2y+3)(6y-5) .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.