#### Answer

$64y^2-9=(8y+3)(8y-3)$

#### 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 $\sqrt9=3$
$64y^2-9=(8y+3)(8y-3)$

Published by
Prentice Hall

ISBN 10:
0133500403

ISBN 13:
978-0-13350-040-0

$64y^2-9=(8y+3)(8y-3)$

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