Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 0 - Section 0.5 - Solving Polynomial Equations - Exercises - Page 30: 40

Answer

$$x = 1,{\text{ }}x = \pm \sqrt 3 $$

Work Step by Step

$$\eqalign{ & {x^3} - {x^2} - 3x + 3 = 0 \cr & {\text{Grouping terms}} \cr & \left( {{x^3} - {x^2}} \right) - \left( {3x - 3} \right) = 0 \cr & {\text{Factoring}} \cr & {x^2}\left( {x - 1} \right) - 3\left( {x - 1} \right) = 0 \cr & \left( {x - 1} \right)\left( {{x^2} - 3} \right) = 0 \cr & {\text{Zero - factor property}} \cr & x - 1 = 0,{\text{ }}{x^2} - 3 = 0 \cr & x = 1,{\text{ }}x = \pm \sqrt 3 \cr} $$
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