Find roots of
$$x^3-(2\sqrt 2+i)x^2 + (4+2i\sqrt 2)x-4i=0$$ knowing that at least one of the roots is from the set $\{i, -i\}$. Write each root in polar form.
I know that since at least one root is from this set, hence ${x}^2 + 1 $ will be a divisor of my initial equation. Usually, I would divide the first one by the second but I came across some difficulties and I don't know what to do.
[EDIT] My assumption was wrong.
On
"I know that since at least one root is from this set, hence $x^2+1$ will be a divisor of my initial equation." This is an incorrect statement—but incorrect for a very interesting and educational reason.
The relevant fact is: if $p(x)$ is a polynomial with coefficients in a field $F$, and $\alpha$ is a root of $p(x)$ (in some extension of $F$), then $p(x)$ is divisible by the minimal polynomial of $\alpha$ over $F$.
You tried to apply this with $\alpha=i$ (say); but the coefficients of the given polynomial are not in $\Bbb R$, so this fact doesn't concern the minimial polynomial $x^2+1$ of $i$ over $\Bbb R$; rather, the coefficients of our $p(x)$ are defined over $\Bbb C$, and so it is the minimal polynomial $x-i$ of $i$ over $\Bbb C$ that is relevant. In other words, we have just (re)learned that $x-i$ divides $p(x)$ (if it is $i$ that is the root of $p(x)$).
Here's an extreme example of the same fallacy: since $\sqrt[4]2$ is a root of $x-\sqrt[4]2$, does it then follow that $x^4-2$ divides $x-\sqrt[4]2$?
Hint. First verify whether $x_0=i$ or $x_0=-i$ is a root ($x^2+1$ is a divisor of the given polynomial iff $x_0=i$ and $x_0=-i$ are roots):
$$(\pm i)^3-(2\sqrt 2+i)(\pm i)^2 + (4+2i\sqrt 2)(\pm i)-4i=0 ?$$ Then divide the cubic polynomial by $x-x_0$. The quotient will be a quadratic polynomial which should be easy to solve.