Answer
$\{-3,-3,0, \frac{1\pm i\sqrt {31}}{4}\}$
Work Step by Step
Step 1. Given $f(x)=2x^5+11x^4+16x^3+15x^2+36x=x(2x^4+11x^3+16x^2+15x+36)$, we can identify one zero $x=0$ and list possible rational zeros for the remaining polynomial as $\pm1,\pm2,\pm4,\pm6,\pm9,\pm12,\pm18,\pm36,\pm\frac{1}{2},\pm\frac{9}{2}$
Step 2. There are 0 sign changes in $f(x)$ indicating that there is no positive real zero.
Step 3. $f(-x)=-2x^5+11x^4-16x^3+15x^2-36x$, there are 4 sign changes in $f(-x)$ indicating that there could be 4, 2, or 0 negative real zero.
Step 4. Use synthetic division to find two real zeros as shown in the figure.
Step 5. The resulting quotient gives $2x^2-x+4=0$ which gives $x=\frac{1\pm\sqrt {1^2-4(2)(4)}}{2(2)}=\frac{1\pm i\sqrt {31}}{4}$
Step 6. The zeros are $\{-3,-3,0, \frac{1\pm i\sqrt {31}}{4}\}$
