Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.8 - Solving Equations by Factoring and Problem Solving - Exercise Set - Page 323: 26

Answer

{-4, 0, 4}

Work Step by Step

1. Rearrange expression to equal zero. $n^{3}$ = 16n $n^{3}$ - 16n = 0 2. Factor the expression. n($n^{2}$ - 16) n(n + 4)(n - 4) 3. Apply the zero factor property. a) n = 0 b) n + 4 = 0 c) n - 4 = 0 4. Solve each linear equation. a) n = 0 b) n = -4 c) n = 4 The solutions are {-4, 0, 4}.
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