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: 26

Answer

$(\sqrt{x}-\sqrt{3})^4=(\sqrt{x}+(-\sqrt{3}))^4=\binom{4}{0}(\sqrt{x})^4(-\sqrt{3})^0+\binom{4}{1}(\sqrt{x})^3(-\sqrt{3})^1+\binom{4}{2}(\sqrt{x})^2(-\sqrt{3})^2+\binom{4}{3}(\sqrt{x})^1(-\sqrt{3})^3+\binom{4}{4}(\sqrt{x})^0(-\sqrt{3})^4=x^2-4(x^{1/2})^3\sqrt{3}+18x-4x^{1/2}3\sqrt{3}+9= x^2-4\sqrt{3}x^{3/2}+18x-12\sqrt{3}x^{1/2}+9$

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 $\sqrt{x}$; $y=-\sqrt{3}$ and $n=4$ $(\sqrt{x}-\sqrt{3})^4=(\sqrt{x}+(-\sqrt{3}))^4=\binom{4}{0}(\sqrt{x})^4(-\sqrt{3})^0+\binom{4}{1}(\sqrt{x})^3(-\sqrt{3})^1+\binom{4}{2}(\sqrt{x})^2(-\sqrt{3})^2+\binom{4}{3}(\sqrt{x})^1(-\sqrt{3})^3+\binom{4}{4}(\sqrt{x})^0(-\sqrt{3})^4=x^2-4(x^{1/2})^3\sqrt{3}+18x-4x^{1/2}3\sqrt{3}+9= x^2-4\sqrt{3}x^{3/2}+18x-12\sqrt{3}x^{1/2}+9$
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