Complex solutions of polynomial question

Question

$2z^3-6z^2+mz+n = 0$ $m, n$ are real and $1+\sqrt{ 2} i$ is a solution. Find $m$ and $n$.

Attempt to solve :

Giving the known theorem $1-\sqrt{2}i$ is also a solution, so we can substitute each time one of the roots and get a system in two unknowns to solve. I find this to be rather complex and not elegant. I wonder, are there any other solutions to this problem, which are perhaps not as tedious?

Answer 1

Hint: $(z-a)(z-b)(z-c)$ has roots $a,b,c$ and if $z_0$ is a complex root, then its conjugate too :).

Answer 2

By Vieta's formulas, the sum of the roots is $ -\frac {6}{2} = 3 $. You know that two of the roots are $ 1 + 2i $ and $ 1- 2i $, so the third root must be $ 1 $. Plugging in $1$ to the polynomial we obtain that $ m + n = 4 $. By Vieta's formulas we also know that the product of the roots is $ -\frac {n}{2} = (1+2i)(1-2i)(1) = 5 $. Therefore, $ n = -10 $ and $ m = 14 $.

Answer 3

Hint: You know that $1+\sqrt{2}i$ and $1+\sqrt{2}i$ are solutions of $$z^3-3z^2+\frac{m}{2}z+\frac{n}{2} = 0$$ -3, the coefficient of $z^2$ is the sum of the solutions, so you easily get the third.