Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 10 - Section 10.5 - Finding Factors of Special Products - Exercise - Page 354: 29



Work Step by Step

This is a difference between two squares, and hence $$b^2-9=(b-3)(b+3).$$ One can check as follows $$(b-3)(b+3)=b^2-3b+3b-9=b^2-9 .$$
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.