Finding the zeroes of the function. I would like to find the zeroes of the following function given that $3-i$ is a zero of $f$:

$f(x) = 2x^4-7x^3-13x^2+68x-30$

Please explain to me how to do this problem. Thanks!

Let $x_1,x_2,x_3,x_4$ be the roots of:
$$f(x) = 2x^4-7x^3-13x^2+68x-30$$
Then by Vieta's formulas:
$$
\begin{cases}
x_1+x_2+x_3+x_4=\frac{7}{2} \\
x_1x_2x_3x_4 = -\frac{30}{2}
\end{cases}
$$
Now you know $x_1=3-i$, and its conjugate must be also a root $x_2=3+i$, so the above reduces to:
$$
\begin{cases}
x_3+x_4=\frac{7}{2} - 6 = -\frac{5}{2} \\
x_3x_4 = -\frac{30}{2} / 10 = -\frac{3}{2}
\end{cases}
$$
Then, again by Vieta's formulas, the remaining roots $x_3,x_4$ must be the roots of:
$$x^2 + \frac{5}{2}x-\frac{3}{2} = 0$$
