Linear factorization of the complex polynomial $z^3 -(5 + i)z^2 +(2+5i)z-10$ Do a linear factorization of the following complex polynomial :
$z^3 -(5 + i)z^2 +(2+5i)z-10 $
Rearranging to $ z(z(z-(5+i))+(2+5i))-10 $ doesn't help
btw: $1, 2, 3, i$ and $(2 +i)$ aren't solutions
 A: Integer roots, if any, are divisors of the zero-degree term, $-10$ in this case.
Because if $ax^3+bx^2+cx=-d$ then $x(ax^2+bx+c)=-d$ which shows that if the polynomial has an integer root this must be a divisor of $d$
This gives the possibilities $\pm 1;\;\pm 2;\;\pm5;\;\pm10$ 
it IS actually quite boring, but often it works. Then you do the short division and get $z^3-(5+i) z^2+(2+5 i) z-10=(x-5)(z^2-i z+2)$
and then 
$$z^3-(5+i) z^2+(2+5 i) z-10=(z-5) (z+i) (z-2 i)$$
A: To find possible real solutions, seperate real-part and imaginary-part. You get 
$$z^3-5z^2+2z-10$$ and $$-z^2+5z$$
The imaginary part factors as $z(-z+5)$ , the real part as $(z-5)(z^2+2)$. This way, it is easy to detect the solution $z=5$
If you substitute $z$ by $ui$ and seperate real and imaginary part, you get $$5u^2-5u-10=5(u-2)(u+1)$$ and $$-u^3+u^2+2u=-(u-2)u(u+1)$$ giving the solutions $2i$ and $-i$
A: $(z + i)\cdot(z - 2i)\cdot(z - 5)$
But why is this a problem? (please see comments below)
A: This a polynomial over the ring of Gaussian integers $\mathbf Z[i]$, which is a P.I.D. with units $\;1,-1,i,-i$.  As a consequence, the rational roots theorem remains valid in this ring, so that a ‘rational root’ of this polynomial is a divisor of $10$.
It happens that $5$ is a root, and we obtain by Horner's scheme the factorisation
$$z^3-(5+i)z^2+(2+5i)z-10=(z-5)(z^2-iz+2). $$
Now the quadratic polynomial $z^2-iz+2$ has the obvious root $-i$, so, as the product of the roots is equal to  $2$, the other root is $2i$, and we have the factorisation
$$z^3-(5+i)z^2+(2+5i)z-10=(z-5)(z+i)(z-2i).$$
