Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Section 11.4 - The Binomial Theorem - Exercise Set - Page 864: 14

Answer

$64x^3-48x^2+12x-1$

Work Step by Step

According to the Binomial Theorem or Binomial expansion, the expansion for $(x+y)$ in the any number of power can be calculated as: $(x+y)^n=\displaystyle \binom{n}{0}x^ny^0+\displaystyle \binom{n}{1}x^{n-1}y^1+........+\displaystyle \binom{n}{n}x^0y^n$ Let's see how this formula works: $(4x-1)^3=\displaystyle \binom{3}{0}(4x)^3(-1)^0+\displaystyle \binom{3}{1}(4x^{2})(-1)^1 +\displaystyle \binom{3}{2}(4x)^1(-1)^2+\displaystyle \binom{3}{3}(4x)^0(-1)^3$ or, $=64x^3+3(16x^2)(-1)+3(4x)(1)+(-1)$ or, $=64x^3-48x^2+12x-1$
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