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: 23

Answer

$(x^2+y^2)^5=\binom{5}{0}(x^2)^5(y^2)^0+\binom{5}{1}(x^2)^4(y^2)^1+\binom{5}{2}(x^2)^3(y^2)^2+\binom{5}{3}(x^2)^2(y^2)^3+\binom{5}{4}(x^2)^1(y^2)^4+\binom{5}{5}(x^2)^0(y^2)^5=x^{10}\times 1+5x^8y^2+10x^6y^4+10x^4y^6+5x^2y^8+y^{10}\times 1=$ $x^{10} 1+5x^8y^2+10x^6y^4+10x^4y^6+5x^2y^8+y^{10}$

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 $x^2$ and y becomes $y^2$; $n=5$ $(x^2+y^2)^5=\binom{5}{0}(x^2)^5(y^2)^0+\binom{5}{1}(x^2)^4(y^2)^1+\binom{5}{2}(x^2)^3(y^2)^2+\binom{5}{3}(x^2)^2(y^2)^3+\binom{5}{4}(x^2)^1(y^2)^4+\binom{5}{5}(x^2)^0(y^2)^5=x^{10}\times 1+5x^8y^2+10x^6y^4+10x^4y^6+5x^2y^8+y^{10}\times 1=$ $x^{10} 1+5x^8y^2+10x^6y^4+10x^4y^6+5x^2y^8+y^{10}$
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