Question 172910
I'm assuming that you want to find the roots? Please post full instructions.



{{{f(x)=x^3-4x}}} Start with the given equation



{{{0=x^3-4x}}} Plug in {{{f(x)=0}}}



{{{0=x(x^2-4)}}} Factor out the GCF "x"



{{{0=x(x+2)(x-2)}}} Factor {{{x^2-4}}} to get {{{(x+2)(x-2)}}} (by using the difference of squares)




Now set each factor equal to zero:



{{{x=0}}}, {{{x+2=0}}} or {{{x-2=0}}}



{{{x=0}}}, {{{x=-2}}} or {{{x=2}}}   Now solve for x in each case




So the roots are 


 {{{x=0}}}, {{{x=-2}}} or {{{x=2}}}