Question 825385
As per the rational zero theorem,
the possible rational zeros of {{{f(x)= x^3-4x^2-11x+2}}} are integers factors of {{{2}}} , meaning
-2, -1, 1, and 2.
 
We can try them all, starting by the easiest ones.
Trying 1 and -1 by substitution is easy enough:
{{{f(1)1-4-11+2=-3-11+2=-14+2=-12}}}
{{{f(-1)=-1-4+11+2=-5+11+2=-6+2=8}}} .
That means that {{{x=1}}} is not a zero of {{{f(x)}}} , and neither is {{{x=-1}}} .
If {{{x=2}}} is a zero of {{{f(x)}}} , {{{x-2}}} must be a factor of {{{f(x)}}} . If {{{x=-2}}} is a zero of {{{f(x)}}} , {{{x-(-2)=x+2}}} must be a factor of {{{f(x)}}} .
We could try -2 and 2 by substitution, but since in the end we will need to divide, we may choose just to try dividing by {{{x-2}}} and by {{{x+2}}} .
We would find that dividing {{{f(x)}}} by {{{x-2}}} leaves a remainder,
but that {{{f(x)}}} divides evenly by {{{x+2}}} , and
{{{f(x)=(x+2)(x^2-6x+1)}}} .
That means that {{{highlight(x=-2)}}} is a zero of {{{f(x)}}} ,
and the remaining zeros are the zeros of {{{x^2-6x+1}}} .
 
We can use find the zeros of {{{x^2-6x+1}}} by solving the quadratic equation
{{{x^2-6x+1=0}}} either by using the quadratic formula or by "completing the square":
{{{x^2-6x+1=0}}}
{{{x^2-6x=-1}}}
{{{x^2-6x+9=-1+9}}}
{{{(x-3)^2=8}}}
So either {{{x-3=sqrt(8)=2sqrt(2)}}} --> {{{highlight(x=3+2sqrt(2))}}} ,
or {{{x-3=-sqrt(8)=-2sqrt(2)}}} --> {{{highlight(x=3-2sqrt(2))}}} .
 
In sum, the zeros of {{{f(x)= x^3-4x^2-11x+2}}} are
{{{highlight(x=-2)}}} , {{{highlight(x=3+2sqrt(2))}}} , and {{{highlight(x=3-2sqrt(2))}}} .