Answer
$(y-2)^5=y^5-10y^4+40y^3-80y^2+80y-32$
Work Step by Step
$(y-2)^5=~_5C_0y^5+~_5C_1y^4(-2)^1+~_5C_2y^3(-2)^2+~_5C_3y^2(-2)^3+~_5C_4y^1(-2)^4+~_5C_5(-2)^5=y^5+5y^4(-2)+10y^3(4)+10y^2(-8)+5y(16)+(-32)=y^5-10y^4+40y^3-80y^2+80y-32$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 11 - 11.5 - The Binomial Theorem - 11.5 Exercises - Page 813: 20
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