Finding the zeroes of the function.

Question

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!

Answer 1

If you know that $3-i$ is a zero, then you know that $3+i$ is also a zero. See this.

This implies that $\left(x^2-6x+10\right)$ is a factor of the original polynomial.
Can you pull it off from here?

Answer 2

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:

$$x^2 + \frac{5}{2}x-\frac{3}{2} = 0$$