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: 27

Answer

{-1, 1, 2}

Work Step by Step

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