<s>[INST] Factorising $(4 + 3i)z^2 + 26iz + (-4+3i)$? Quadratic formula attempt included. I'm trying to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$.
I tried to use the quadratic formula to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$, where $b = 26i$, $a = (4 + 3i)$, $c = (-4+3i)$. This gives us the roots $z = \dfrac{-4i - 12}{25}$ and $z =-4i - 3$.
But $\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right) \not= (4 + 3i)z^2 + 26iz + (-4+3i)$. 
So I did what we do when using the quadratic formula with real numbers and have that $a \not= 0, 1$: $(4 + 3i)\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right)$. But, as far as I can tell, $(4 + 3i)\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right) \not= (4 + 3i)z^2 + 26iz + (-4+3i)$.
So what is the correct way to go about this?
I would greatly appreciate it if people could please take the time to explain this.
 [/INST] The solutions of the equation $(4+3i)z^2+26iz+(3i-4)=0$ are $z=-3-4i$ and $z=\dfrac{-3-4i}{25}$. Try using the Factor Theorem now, it should work. 
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