## College Algebra (10th Edition)

Published by Pearson

# Chapter 9 - Section 9.5 - The Binomial Theorem - 9.5 Assess Your Understanding - Page 676: 28

#### Answer

$(ax-by)^4=(ax+(-by))^4=\binom{4}{0}(ax)^4(-by)^0+\binom{4}{1}(ax)^3(-by)^1+\binom{4}{2}(ax)^2(-by)^2+\binom{4}{3}(ax)^1(-by)^3+\binom{4}{4}(ax)^0(-by)^4=a^4x^4-4a^3bx^3y+6a^2b^2x^2y^2-4ab^3xy^3+b^4y^4$

#### 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 $ax$ , y becomes $-by$, $n=4$ $(ax-by)^4=(ax+(-by))^4=\binom{4}{0}(ax)^4(-by)^0+\binom{4}{1}(ax)^3(-by)^1+\binom{4}{2}(ax)^2(-by)^2+\binom{4}{3}(ax)^1(-by)^3+\binom{4}{4}(ax)^0(-by)^4=a^4x^4-4a^3bx^3y+6a^2b^2x^2y^2-4ab^3xy^3+b^4y^4$

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