Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.3 - Algebraic Expressions - 1.3 Exercises - Page 34: 43

Answer

$ y^3 + 6y^2 + 12y + 8$

Work Step by Step

$Multiply$ $the$ $algebraic$ $expressions$ $using$ $a$ $Special$ $Product$ $Formula$ $and$ $simplify:$ $(y+2)^3$ Use the Cube of a Sum Product Formula: $(A+B)^2 = A^3 + 3A^2B + 3AB^2 + B^3$ $(y+2)^3$ = $y^3$ + $(3\times y^2 \times 2)$ + $(3\times y \times 2^2)$ + $2^3$ $= y^3 + 6y^2 + 12y + 8$
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