Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.4 - Adding, Subtracting, and Multiplying Radical Expressions - Exercise Set: 63

Answer

$(\sqrt{6}-4\sqrt{2})(3\sqrt{6}+1)=18+\sqrt{6}-24\sqrt{3}-4\sqrt{2}$

Work Step by Step

$(\sqrt{6}-4\sqrt{2})(3\sqrt{6}+1)$ Evaluate the product: $(\sqrt{6}-4\sqrt{2})(3\sqrt{6}+1)=3\sqrt{6^{2}}+\sqrt{6}-12\sqrt{12}-4\sqrt{2}=...$ $...=3(6)+\sqrt{6}-12\sqrt{12}-4\sqrt{2}=...$ $...=18+\sqrt{6}-12\sqrt{12}-4\sqrt{2}=...$ Rewrite this expression as $18+\sqrt{6}-12\sqrt{4\cdot3}-4\sqrt{2}$ and simplify: $...=18+\sqrt{6}-12\sqrt{4\cdot3}-4\sqrt{2}=...$ $...=18+\sqrt{6}-12(2)\sqrt{3}-4\sqrt{2}=$ $...=18+\sqrt{6}-24\sqrt{3}-4\sqrt{2}$
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