Answer
$x=\frac{5\pm \sqrt {37}}{6}$
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 $x=3$:
$f(3)=3(3)^4-8(3)^3-14(3)^2+31(3)+6=0$ and
We check $x=-2$:
$f(-2)=3(-2)^4-8(-2)^3-14(-2)^2+31(-2)+6=0$
thus, it is indeed a zero.
Then, we can write:
$f(x)=(x+2)(x-3)(3x^2-5x-1)$, solving for the trinomial using quadratic formula for the quadratic equation $ax^2+bx+c$, $x=\frac{-b\pm \sqrt {b^2-4ac}}{2a}$.
for our trinomial, $3x^2-5x-1$, $x=\frac{5\pm \sqrt {(-5)^2-4\times3 \times (-1)}}{2\times 3}= \frac{5\pm \sqrt {37}}{6}$
Thus the other zeros are:
$x=\frac{5\pm \sqrt {37}}{6}$
