## Elementary Algebra

{$-2,0,2$}
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 {$-2,0,2$}.