Answer
$\{-4,-\frac{1}{2},2,2\}$
Work Step by Step
Step 1. Given $f(x)=2x^4+x^3-24x^2+20x+16$, list possible rational zeros as $\frac{p}{q}=\pm1,\pm2,\pm4,\pm8,\pm16,\pm\frac{1}{2}$
Step 2. Use synthetic division as shown in the figure to find zero(s) $x=2,-4$.
Step 3. Use the quotient and solve $2x^2-3x-2=0$ or $(x-2)(2x+1)=0$, thus $x=-\frac{1}{2}, 2$.
Step 4. Thus the real zeros are $\{-4,-\frac{1}{2},2,2\}$
