<s>[INST] Find all solutions of the equation $z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10=0$ Find all solutions of the equation:
$z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10=0$
I don't have any  other idea than to guess them.
 [/INST] The peculiar form of the polynomial, with the same $\,\sqrt{2}\,$ present in the imaginary parts of the coefficients, only, suggests splitting it into two terms for closer inspection:
$$
\begin{align}
P(z) &= z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10 \\
&= (z^4-2z^3+3z^2+4z-10) + i \,\sqrt{2}\,(2z^3 - 4 z^2+10z) \\
&= (z^4-2z^3+3z^2+4z-10) + i \,2\sqrt{2}\,z(z^2 - 2 z+5) \\
\end{align}
$$
The quadratic factor $\,z^2 - 2 z+5\,$ in the second term further suggests checking whether the first term maybe has it as a factor as well, and it turns out that it does, indeed:
$$
z^4-2z^3+3z^2+4z-10 = (z^2-2)(z^2 - 2 z+5)
$$
From here on, the polynomial factors, and the rest follows by routine calculations:
$$
P(z) = (z^2 + i\,2\,\sqrt{2}z- 2)(z^2-2z+5)
$$
Disclaimer: the shortcut above only works because the polynomial was specifically set up the way it was. Such "luck" virtually never happens in "real life" outside homework/contest problems.
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