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: 34



Work Step by Step

This is a difference between two squares, and hence $$ 16b^2-81=(4b-9)(4b+9) .$$ One can check as follows $$ (4b-9)(4b+9)=16b^2-36b+36b-81=16b^2-81 .$$
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