#### Answer

$\sqrt {6}+\sqrt 2-4\sqrt 3-4$

#### Work Step by Step

$(\sqrt 2-4)(\sqrt 3+1)$
=$\sqrt 2(\sqrt 3+1)-4(\sqrt 3+1)$
=$\sqrt 2\sqrt 3+\sqrt 2-4\sqrt 3-4$
=$\sqrt {2\times3}+\sqrt 2-4\sqrt 3-4$
=$\sqrt {6}+\sqrt 2-4\sqrt 3-4$

Published by
Pearson

ISBN 10:
0321726391

ISBN 13:
978-0-32172-639-1

$\sqrt {6}+\sqrt 2-4\sqrt 3-4$

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