Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Mid-Chapter Check Point - Page 71: 34



Work Step by Step

The first three terms form a perfect square, $(a -b)^{2}=a^{2}-2ab+b^{2}$ a=x and b=3 the last term is a square of $7y$ $=( x-6)^{2}-(7y)^{2}$ now we have a difference of squares, $(a+b)(a-b)=a^{2}-b^{2}$ =$[(x-3)+7y]\cdot[(x-3)-7y]$ = $(x-3+7y)(x-3-7y)$
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