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: 18

Answer

$(x-1)^5=(x+(-1))^5=\binom{5}{0}x^5(-1)^0+\binom{5}{1}x^{4}(-1)^1+\binom{5}{2}x^{3}(-1)^2+\binom{5}{3}x^{2}(-1)^3+\binom{5}{4}x^{1}(-1)^{4}+\binom{5}{5}x^0(-1)^5=x^5-5x^4+10x^3-10x^2+5x-1$

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, $y=-1; n=5$ $(x-1)^5=(x+(-1))^5=\binom{5}{0}x^5(-1)^0+\binom{5}{1}x^{4}(-1)^1+\binom{5}{2}x^{3}(-1)^2+\binom{5}{3}x^{2}(-1)^3+\binom{5}{4}x^{1}(-1)^{4}+\binom{5}{5}x^0(-1)^5=x^5-5x^4+10x^3-10x^2+5x-1$
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.