I was reading through something which progressed from the line:
$z^3 + 2(2 + i)z^2 + (7 + 10i)z+ 6(1 + 2i) = 0$
To this line, factorised:
$(z-2) (z^2 - 2(1 + i)z + 3(1+2i) ) = 0$
With no further information. Any advice as to how this step can be performed would be appreciated, as my own attempts don't appear to be working.
On
Let $R = \mathbb{Z}[i]$, the ring of Gaussian integers.
Let $f = z^3 + 2(2 + i)z^2 + (7 + 10i)z+ 6(1 + 2i)$.
Regarding $f$ as an element of $R[z]$, then since $R[z]$ is a UFD, and since $f$ is cubic and monic, it follows that $f$ is reducible in $R[z]$ if and only if $f$ has a root $r \in R$.
Then $r$ must be a factor the constant term $6(1+2i)$.
Noting that $1 + 2i$ is prime in $R$, a necessary condition is: $$r \in \{u,2u,3u,6u,u(1+2i),2u(1+2i),3u(1+2i),6u(1+2i)\}$$ where $u$ is a unit of $R$, i.e., $u \in \{\pm 1,\pm i\}$.
Testing, we find $f(-2) = 0$, hence $z+2$ is a factor of $f$.
By Vieta's formulas, the sum of the roots of $f$ is $-2(2+i)$, and the product of the roots of $f$ is $-6(1+2i)$.
Hence, removing the known root $-2$, the sum of the other two roots is $$-2(2+i)-(-2) =-2(1+i)$$ and the product of the other two roots is $$\frac{-6(1+2i)}{-2}=3(1+2i)$$
Then by Vieta's formulas, the other two roots are roots of the quadratic polynomial $$z^2+2(1+i)z+3(1+2i)$$ It follows that $$f = (z+2)(z^2+2(1+i)z+3(1+2i))$$
$$z^3 + 2(2 + i)z^2 + (7 + 10i)z+ 6(1 + 2i)=(z+2)(z+3-i)(z+1+3i)$$