Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 6 - Factoring, Solving Equations, and Problem Solving - Chapter 6 Review Problem Set - Page 274: 37



Work Step by Step

Since $2x$ is common to both the terms of the equation, we take it out as a common factor: $2x^{3}-8x=0$ $2x(x^{2}-4)=0$ We simplify the expression further using the rule $a^{2}-b^{2}=(a+b)(a-b)$: $2x(x^{2}-4)=0$ $2x(x^{2}-2^{2})=0$ $2x(x+2)(x-2)=0$ Now, we equate all the factors to zero to solve the equation: $2x(x+2)(x-2)=0$ $2x=0$ or $(x+2)=0$ or $(x-2)=0$ $x=0$ or $x=-2$ or $x=2$ Therefore, the solution set is {$-2,0,2$}.
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