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 - 10-3 Operations With Radical Expressions - Practice and Problem-Solving Exercises - Page 629: 24


The answer is $12-8\sqrt3$.

Work Step by Step

We can simplify each product as follows: $=-\sqrt12(4-2\sqrt3)$ $=-\sqrt12*4+2*\sqrt3*\sqrt12$ $=-\sqrt4*\sqrt3*4+2*\sqrt3*\sqrt4*\sqrt3$ $=-2*\sqrt3*4+2*\sqrt3*2*\sqrt3$ $=-8*\sqrt3+4*3$ $=-8\sqrt3+12$ $=12-8\sqrt3$.
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