Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 10 - Radical Expressions and Equations - 10-2 Simplifying Radicals - Practice and Problem-Solving Exercises - Page 623: 29

Answer

$16y^3$

Work Step by Step

Simpifying $\sqrt 2y \times \sqrt 128y^5$ $ \sqrt 2 \times \sqrt 128 \times\sqrt y \times \sqrt y^5$ = $\sqrt 256y^6$ Finding the perfect square factor of $ 256y^6$ $256y^6 = 4 \times 64 \times y^6$ = $2^2 \times8^2 \times y^6$ = $(2 \times8 \times y^3)^2$ = $(16y^3)^2$ $\sqrt 256y^6 = \sqrt (16y^3)^2$ =$\sqrt16y^3$
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