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: 43

Answer

$$x = - 2,{\text{ }}x = - 1,{\text{ }}x = 3,{\text{ }}x = 2$$

Work Step by Step

$$\eqalign{ & \left( {{x^2} + 3x + 2} \right)\left( {{x^2} - 5x + 6} \right) = 0 \cr & {\text{Factoring each trinomial}} \cr & {x^2} + 3x + 2 = \left( {x + 2} \right)\left( {x + 1} \right) \cr & {x^2} - 5x + 6 = \left( {x - 3} \right)\left( {x - 2} \right) \cr & {\text{Therefore}} \cr & \left( {x + 2} \right)\left( {x + 1} \right)\left( {x - 3} \right)\left( {x - 2} \right) = 0 \cr & {\text{Zero - factor property}} \cr & x + 2 = 0,{\text{ }}x + 1 = 0,{\text{ }}x - 3 = 0,{\text{ }}x - 2 = 0 \cr & {\text{Solving all the linear equations}} \cr & x = - 2,{\text{ }}x = - 1,{\text{ }}x = 3,{\text{ }}x = 2 \cr} $$
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