Answer
$f(x)=2(x+2)(x+\sqrt{5})(x-\sqrt{5})$
Zeros: $-\sqrt{5},\ \ -2,\ \ \sqrt{5},$
all with multiplicity 1.
Work Step by Step
$f(x)=3x^{3}+6x^{2}-15x-30=3(x^{3}+2x^{2}-5x-10)$
$f(x)$ has at most 3 real zeros, as the degree is 3.
Rational zeros: $\displaystyle \frac{p}{q},$ in lowest terms,
$p$ must be a factor of $-10,\qquad \pm 1,\pm 2,\pm 5,\pm 10$
and $q$ must be a factor of $ 2,\qquad \pm 1$
$\displaystyle \frac{p}{q}=\pm 1,\pm 2,\pm 5,\pm 10$
Testing with synthetic division, try $x+2$
$\left.\begin{array}{l}
-2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 &2 &-5 &-10 \\\hline
&-2 &0 &+10 \\\hline
1&0 &-5& |\ \ 0 \end{array}$
$f(x)=2(x+2)(x^{2}-5)$
Solving $x^{2}-5=0,$
$x=\pm\sqrt{5}$, we have the other two zeros.
$f(x)=2(x+2)(x+\sqrt{5})(x-\sqrt{5})$
Zeros: $-\sqrt{5},\ \ -2,\ \ \sqrt{5},$
all with multiplicity 1.