The possible rational zeros are \(\pm1\), \(\pm5\), \(\pm\frac{1}{2}\), \(\pm\frac{5}{2}\). Applying abbreviated synthetic division, we find
| \(2\) | \(3\) | \(8\) | \(-5\) | |
| \(1\) | \(2\) | \(5\) | \(13\) | \(8\) |
| \(5\) | \(2\) | \(13\) | \(73\) | \(360\) |
| \(-1\) | \(2\) | \(1\) | \(7\) | \(-12\) |
| \(-5\) | \(2\) | \(-7\) | \(43\) | \(-220\) |
| \(\frac{1}{2}\) | \(2\) | \(4\) | \(10\) | \(0\) |
Since \(\frac{1}{2}\) gives a remainder of \(0\), \(\frac{1}{2}\) is a zero of the function. Using the result of the last row we see that the polynomial factors as \[ \left(x-\frac{1}{2}\right)(2x^2+4x+10)=2\left(x-\frac{1}{2}\right)(x^2+2x+5)=(2x-1)(x^2+2x+5) \]
Thus, the complex zeros are the zeros of \(x^2+2x+5\).
Setting \(x^2+2x+5=0\) and solving for \(x\) using the quadratic formula we find the complex zeros:
\[ x=\dfrac{-2\pm\sqrt{4-20}}{2}=-1\pm2\,i \]