Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set - Page 309: 38



Work Step by Step

Factoring the $GCF= 2 $ results to $ 2(27y^3-64) $. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$, or the factoring of the sum/difference of two cubes, then, \begin{array}{l} 2(27y^3-64) \\= 2[(3y)+(-4)][(3y)^2-(3y)(-4)+(-4)^2) \\= 2(3y-4)(9y^2+12y+16) .\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.