Question 192286
First, let's factor {{{3x^3-x^2-18x+6}}}



{{{3x^3-x^2-18x+6}}} Start with the given expression



{{{(3x^3-x^2)+(-18x+6)}}} Group like terms



{{{x^2(3x-1)-6(3x-1)}}} Factor out the GCF {{{x^2}}} out of the first group. Factor out the GCF {{{-6}}} out of the second group



{{{(x^2-6)(3x-1)}}} Since we have the common term {{{3x-1}}}, we can combine like terms



So {{{3x^3-x^2-18x+6}}} factors to {{{(x^2-6)(3x-1)}}}



This means that {{{f(x)= 3x^3-x^2-18x+6}}} factors to {{{f(x)=(x^2-6)(3x-1)}}}



To find the zeros, set the right side equal to zero:


{{{(x^2-6)(3x-1)=0}}}



{{{x^2-6=0}}} or {{{3x-1=0}}} Use the zero product property



{{{x=sqrt(6)}}}, {{{x=-sqrt(6)}}} or {{{x=1/3}}} Solve for "x" in each case



So the zeros are {{{x=sqrt(6)}}}, {{{x=-sqrt(6)}}} or {{{x=1/3}}}