Answer
$\{-5, 3, \pm2i\sqrt 3 \}$
Work Step by Step
Step 1. Given $f(x)=x^4+2x^3-3x^2+24x-180$, we can list possible rational zeros as $\pm1,\pm2,\pm3,\pm4,\pm5,\pm6,\pm9,\pm10,\pm12,\pm18,\pm20,...$
Step 2. There are 3 sign changes in $f(x)$ indicating that there could be 3 or 1 positive real zeros.
Step 3. $f(-x)=x^4-2x^3-3x^2-24x-180$, there is 1 sign change in $f(-x)$ indicating that there must be 1 negative real zero.
Step 4. Use synthetic division to find two real zeros as shown in the figure.
Step 5. The resulting quotient gives $x^2+12=0$ which gives $x=\pm2i\sqrt 3$
Step 6. The zeros are $\{-5, 3, \pm2i\sqrt 3 \}$