polynomial with complex roots 
As I understans - this polynomial has 4 different roots. We are given $z_1=3-2i$, so $z_2=3+2i$. So we already have 2 of 4 roots. I expressed the polynomial as $(x^2+2)(x^2-6x+13)$. I solved each of these equations separately and got $x_{1,2}=\pm i$ for $(x^2+2)$ and $x_{3,4}=\pm 2i$ for $(x^2-6x+13)$
So there are 6 roots in total. What am I doing wrong here?
Thanks in advance.

I think you are miscalculating the roots of the polynomial.
Indeed, since $3 - 2i$ is a root and the coefficients of $p(z)$ are real, $3 + 2i$ is also a root.
Taking advantage of such fact, we can conclude that $z^{2} - 6z + 13$ is a factor of $p(z)$.
Hence we have that
\begin{align*}
z^{4} - 6z^{3} + 15z^{2} - 12z + 26 & = (z^{4} - 6z^{3} + 13z^{2}) + (2z^{2} - 12z + 26)\\\\
& = z^{2}(z^{2} - 6z + 13) + 2(z^{2} - 6z + 13)\\\\
& = (z^{2} + 2)(z^{2} - 6z + 13)
\end{align*}
whence it results the other roots are given by $\pm i\sqrt{2}$.
Hopefully this helps!
