Question 147703
{{{f (x) = (x^2-9)^2}}} Start with the given equation.



{{{0 = (x^2-9)^2}}} Plug in {{{f(x)=0}}}



{{{0 = (x^2-9)(x^2-9)}}} Expand.



{{{x^2-9=0}}} or {{{x^2-9=0}}} Set each factor equal to zero.



Let's solve the first equation {{{x^2-9=0}}}



{{{x^2-9=0}}} Start with the given equation.



{{{(x+3)(x-3)=0}}} Factor the left side.



{{{x+3=0}}} or {{{x-3=0}}} Set each factor equal to zero.



{{{x=-3}}} or {{{x=3}}} Solve for x.



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If we solve the second equation {{{x^2-9=0}}}, we will get the same answers {{{x=-3}}} or {{{x=3}}}. So that means that both solutions occur twice. So the solutions are



{{{x=-3}}} (with multiplicity of 2) or {{{x=3}}} (with multiplicity of 2)