Answer
$\{-\frac{1}{5},1\pm i\sqrt 5\}$
Work Step by Step
Step 1. Given $f(x)=5x^3-9x^2+28x+6$, we can list possible rational zeros as $\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{3}{5},\pm\frac{6}{5}$
Step 2. There are 2 sign changes in $f(x)$ indicating that there could be 2 or 0 positive real zeros.
Step 3. $f(-x)=-5x^3-9x^2-28x+6$, there is 1 sign change in $f(-x)$ indicating that there will be 1 negative real zero.
Step 4. Use synthetic division to find one real zeros as shown in the figure.
Step 5. The resulting quotient gives $5x^2-10x+30=0$ or $x^2-2x+6=0$ which gives $x=\frac{2\pm\sqrt {2^2-4(6)}}{2}=1\pm i\sqrt 5$
Step 6. The zeros are $\{-\frac{1}{5},1\pm i\sqrt 5\}$
