Answer
$\{-2, \pm \sqrt 5 \}$, $f(x)=3(x+2)(x+\sqrt 5)(x-\sqrt 5)$
Work Step by Step
Step 1. Given $f(x)=3x^3+6x^2-15x-30$, list possible rational zeros as $\frac{p}{q}=\pm1,\pm2,\pm3,\pm5,\pm6,\pm10,\pm15,\pm30,\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{5}{3},\pm\frac{10}{3}$
Step 2. Use synthetic division as shown in the figure to find one zero $x=-2$.
Step 3. Use the quotient to solve $3x^2-15=0$ or $x^2=5$, thus $x=\pm \sqrt 5$
Step 4. Thus the real zeros are $\{-2, \pm \sqrt 5 \}$ and we can factor the function as $f(x)=3(x+2)(x+\sqrt 5)(x-\sqrt 5)$
