# 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, $(4 + 3i)\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right) \not= (4 + 3i)z^2 + 26iz + (-4+3i)$.

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.
Another way, let $a = 4+3i$, then the equation is $az^2 + (a \overline a+1)iz - \overline{a}=0$.
$$\implies (az)^2 + (a \overline ai+i)(az) - a\overline{a}=0$$
Knowing Vieta, it is easy to observe the roots are $az \in \{ -a\overline ai, -i\}$, which implies $z \in \{-\overline ai, -\dfrac{i}a \} = \{-3-4i, -\frac3{25}-\frac4{25}i\}$.