Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Mid-Chapter Check Point - Page 541: 22

Answer

$8x^2y^3\sqrt {y}$.

Work Step by Step

The given expression is $=\sqrt {32xy^2}\cdot \sqrt {2x^3y^5}$ Use the product rule. $=\sqrt {32xy^2\cdot 2x^3y^5}$ Multiply using power rule. $=\sqrt {64x^4y^7}$ Factor the radicands as a perfect square. $=\sqrt {8^2(x^2)^2(y^3)^2y}$ Simplify. $=8x^2y^3\sqrt {y}$.
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