Answer
the other zeros are:
$x=\frac{-1\pm \sqrt {21}}{2}$
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 $x=3$:
$f(3)=3^3-(3)^2-11(3)+15=0$
thus, it is indeed a zero.
Then, we can write:
$f(x)=(x-3)(x^2+x-5)$, 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, $x^2+x-5$, $x=\frac{-1\pm \sqrt {1^2-4\times1 \times -5}}{2\times 1}= \frac{-1\pm \sqrt {21}}{2}$
Thus the other zeros are:
$x=\frac{-1\pm \sqrt {21}}{2}$
