Factoring $(1+i)z^2 - 3z +2 - i$

Question

I'm not sure how to factor $(1+i)z^2 - 3z +2 - i$. I was thinking of writing this as: $$c(z - \alpha)(z -\beta)$$ and solving for $\alpha,\beta$ however that yields some very nasty results. Could someone give me a starting step?

Answer 1

HINT: $$(1+i)z^2 - 3z +2 - i=(z^2-3z+2)+(z^2-1)i=\color{red}{(z-1)}(z-2)+\color{red}{(z-1)}(z+1)i$$

Answer 2

Hint:  by inspection $\,z=1\,$ is a root, so one of the factors is $\,z-1\,$.

Answer 3

Since $1$ is a root we obtain: $$(1+i)z^2-3z+2-i=(z-1)((1+i)z-2+i)=(1+i)(z-1)\left(z-\frac{2-i}{1+i}\right)=$$ $$=(1+i)(z-1)\left(z-\frac{2-2i-i-1}{2}\right)=(1+i)(z-1)\left(z-\frac{1}{2}+\frac{3}{2}i\right)$$