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: 7



Work Step by Step

$64x^{6}$ $=(4)(4)(4)(x)(x)(x)(x)(x)(x)$ $=(4)(4)(4)(x^{2})(x^{2})(x^{2})$ $=(4x^{2})^{3}$
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