Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.5 The Binomial Theorem - 11.5 Assess Your Understanding - Page 857: 25

Answer

$x^3+6\sqrt 2x^{5/2}+30x^2+40\sqrt 2x^{3/2}+60x+24\sqrt 2x^{1/2}+8$

Work Step by Step

Based on the Binomial Theorem, we have $(\sqrt x+\sqrt 2)^6=(\sqrt x)^6+6(\sqrt 2)(\sqrt x)^5+15(\sqrt 2)^2(\sqrt x)^4+20(\sqrt 2)^3(\sqrt x)^3+15(\sqrt 2)^4(\sqrt x)^2+6(\sqrt 2)^5(\sqrt x)+(\sqrt 2)^6=x^3+6\sqrt 2x^{5/2}+30x^2+40\sqrt 2x^{3/2}+60x+24\sqrt 2x^{1/2}+8$
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