Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.4 - Polynomials - Exercise Set - Page 56: 51

Answer

$x^3+3x^2+3x+1$

Work Step by Step

RECALL: $(a+b)^3=a^3++3a^2b+ 3ab^2+b^3$ Use the formula above with $a=x$ and $b=1$ to obtain: $=x^3+3(x^2)(1)+3(x)(1^2)+1^3 \\=x^3+3x^2+3x+1$
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