Answer
$\{-\frac{3}{4},\pm i\sqrt 2\}$
Work Step by Step
Step 1. Given $f(x)=4x^3+3x^2+8x+6$, we can list possible rational zeros as $\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{4},\pm\frac{3}{4}$
Step 2. There are 0 sign changes in $f(x)$ indicating that there will be 0 positive real zeros.
Step 3. $f(-x)=-4x^3+3x^2-8x+6$, there is 3 sign changes in $f(-x)$ indicating that there will be 3 or 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 $4x^2+8=0$ or $x^2+2=0$ which gives $x=\pm i\sqrt 2$
Step 6. The zeros are $\{-\frac{3}{4},\pm i\sqrt 2\}$
