Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.7 - Factoring by Special Products - Vocabulary, Readiness & Video Check - Page 309: 4

Answer

$(7y^{3})^{2}$

Work Step by Step

$49y^{6}$ $=(7)(7)(y)(y)(y)(y)(y)(y)$ $=(7)(7)(y^{3})(y^{3})$ $=(7y^{3})^{2}$
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