Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - 10-2 Simplifying Radicals - Mixed Review - Page 612: 86



Work Step by Step

Remember that the product of the sum and the difference of the same two terms is the difference of their squares. Since $\sqrt64y^2=8y$ and $\sqrt9=3$ $64y^2-9=(8y+3)(8y-3)$
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