I asked a question about how to solve: $$z^4−2z^3+9z^2−14z+14 = 0$$
When all you know is that there is a root with the real part of 1. I was given great answers and you can find the question here.
(If you feel that this question is redundant or that I should rather update my old question, please feel free to delete this one).
After working some more with this question I think I have found out that the particular solution that I should have been using is the following:
I know that there must be a root: $1 + yi$ and its complex conjugate $1 -yi$.
$z-(1+yi)$ and $z + (1-yi)$ must therefore both be factors to the equation above.
There should therefore be a factor: $(z-1-yi)(z - 1+yi)=z^2-2z+(1+y^2)$.
If I divide the polynomial: $$(z^4 - 2z^3 + 9z^2 - 14z + 14)$$
With: $$(z^2-2z+(1+y^2))$$ I get the quotient: $$z^2 + (8-y^2)$$ And the remainder: $$2z(8-y^2)+6-7y^2+y^4 \longleftrightarrow 2z(8-y^2)+6+(-7+y^2)y^2$$
My strategy: when: $2z(8-y^2)+6+(-7+y^2)y^2 = 0$. I will have the possible values of $y$ and the rest of the roots can be found when the quotient $z^2 + (8-y^2) = 0$.
First of all, is this a viable strategy for solving this problem?
Secondly how do I procede and solve for when $2z(8-y^2)+6+(-7+y^2)y^2 = 0$ ?