Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.5 The Binomial Theorem - 12.5 Assess Your Understanding - Page 836: 26

Answer

$x^2-4\sqrt 3x^{3/2}+18x-12\sqrt 3x^{1/2}+9$

Work Step by Step

We are given the expression: $(\sqrt x-\sqrt 3)^4$ Use the Binomial Theorem to expand the expression: $(\sqrt x-\sqrt 3)^4=\sum_{k=0}^4 \binom{4}{k}(\sqrt x)^{4-k}(-\sqrt 3)^k$ $=\binom{4}{0}(\sqrt x)^4(-\sqrt 3)^0+\binom{4}{1}(\sqrt x)^3(-\sqrt 3)^1+\binom{4}{2}(\sqrt x)^2(-\sqrt 3)^2+\binom{4}{3}(\sqrt x)^1(-\sqrt 3)^3+\binom{4}{4}(\sqrt x)^0(-\sqrt 3)^4$ $=x^2+4x^{3/2}(-\sqrt 3)+6x^1(3)+4x^{1/2}(-3\sqrt 3)+9$ $=x^2-4\sqrt 3x^{3/2}+18x-12\sqrt 3x^{1/2}+9$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.