College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 9 - Section 9.5 - The Binomial Theorem - 9.5 Assess Your Understanding: 25

Answer

$(\sqrt{x}+\sqrt{2})^6=\binom{6}{0}(\sqrt{x})^6(\sqrt{2})^0+\binom{6}{1}(\sqrt{x})^5(\sqrt{2})^1+\binom{6}{2}(\sqrt{x})^4(\sqrt{2})^2+\binom{6}{3}(\sqrt{x})^3(\sqrt{2})^3+\binom{6}{4}(\sqrt{x})^2(\sqrt{2})^4+\binom{6}{5}(\sqrt{x})^1(\sqrt{2})^5+\binom{6}{6}(\sqrt{x})^0(\sqrt{2})^6=x^3+6\sqrt{2}x^{\frac{5}{2}}+30x^2+40\sqrt{2}x^{\frac{3}{2}}+60x+24\sqrt{2}\sqrt{x}+8$

Work Step by Step

The binomial theorem expands an algebraic expression in the form of: $(x+y)^n=\binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+\binom{n}{2}x^{n-2}y^2+...+\binom{n}{i}x^{n-i}y^i+...+\binom{n}{n-1}x^{1}y^{n-1}+\binom{n}{n}x^0y^n=\sum_{i=0}^{n}\binom{n}{i}x^{n-i}y^i$ Here, x becomes $\sqrt{x}$ , $y=\sqrt{2}$, $n=6$ $(\sqrt{x}+\sqrt{2})^6=\binom{6}{0}(\sqrt{x})^6(\sqrt{2})^0+\binom{6}{1}(\sqrt{x})^5(\sqrt{2})^1+\binom{6}{2}(\sqrt{x})^4(\sqrt{2})^2+\binom{6}{3}(\sqrt{x})^3(\sqrt{2})^3+\binom{6}{4}(\sqrt{x})^2(\sqrt{2})^4+\binom{6}{5}(\sqrt{x})^1(\sqrt{2})^5+\binom{6}{6}(\sqrt{x})^0(\sqrt{2})^6=x^3+6\sqrt{2}x^{\frac{5}{2}}+30x^2+40\sqrt{2}x^{\frac{3}{2}}+60x+24\sqrt{2}\sqrt{x}+8$
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