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 - Test - Page 578: 15

Answer

$12\sqrt2-\sqrt{15}$

Work Step by Step

Simplify. $\sqrt 3(4\sqrt 6-\sqrt5)$ Therefore, $\sqrt 3(4\sqrt 6-\sqrt5)=4\sqrt 3\sqrt 6-\sqrt 3\sqrt5$ or, $=4\sqrt{18}-\sqrt{15}$ Thus, the given radical term can be written as: $12\sqrt2-\sqrt{15}$
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