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: 2

Answer

$(2z)^{2}$

Work Step by Step

$4z^{2}$ $=(2)(2)(z)(z)$ $=(2z)^{2}$
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