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: 20

Answer

$(x+3)^5=\binom{5}{0}x^53^0+\binom{5}{1}x^{4}3^1+\binom{5}{2}x^{3}3^2+\binom{5}{3}x^{2}3^3+\binom{5}{4}x^{1}3^{4}+\binom{5}{5}x^03^5=x^5+15x^4+90x^3+270x^2+405x+243$

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, $y=3; n=5$ $(x+3)^5=\binom{5}{0}x^53^0+\binom{5}{1}x^{4}3^1+\binom{5}{2}x^{3}3^2+\binom{5}{3}x^{2}3^3+\binom{5}{4}x^{1}3^{4}+\binom{5}{5}x^03^5=x^5+15x^4+90x^3+270x^2+405x+243$
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