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: 26

Answer

$(1-10y)(1+10y)$

Work Step by Step

This is a difference between two squares, and hence $$1-100y^2=(1-10y)(1+10y).$$ One can check as follows $$(1-10y)(1+10y)=1-10y+10y+100y^2=1-100y^2 .$$
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