Answer
$\{-1, 3\}$
Work Step by Step
The factor theorem says that if $f(c)=0$, then $(x-c)$ is a factor of $f(x)$ and if $(x-c)$ is a factor of $f(x)$, then $f(c)=0$.
We check $x=\frac{1}{3}$:
$f(3)=3(\frac{1}{3})^4-(\frac{1}{3})^3-21(\frac{1}{3})^2-11(\frac{1}{3})+6=0$ and
We check $x=-2$:
$f(-2)=3(-2)^4-(-2)^3-21(-2)^2-11(-2)+6=0$
thus, it is indeed a zero.
Then, we can write:
$f(x)=(x+2)(x-\frac{1}{3})(3x^2-6x-9)$.
Factorize the trinomial factor $(3x^2-6x-9)$
(find two factors of $3(-9)=-27$ whose sum is $-6$):
($-9$ and $+3$).
$3x^2-6x-9=3x^2+3x-9x-9$ ...factor in pairs...
$=3x(x+1)-9(x+1)=(3x-9)(x+1)$.
$f(x)=(x+2)(x-\frac{1}{3})(3x-9)(x+1)$.
Thus the other zeros are:
$x=\{-1, 3\}$
