# Help with particular solution to solving $z^4 - 2z^3 + 9z^2 - 14z + 14 = 0$.

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 $2z(8-y^2)+6+(-7+y^2)y^2 = 0$ ?

Thank you!

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Where it is written "There should therefore be a root..." , I think it should be "there should therefore be a factor..." –  DonAntonio Feb 25 at 12:42
Thank you for pointing that out @DonAntonio –  Lukas Arvidsson Feb 25 at 12:43
Right as far as it goes. Just that you made a mistake: The remainder is $(2 - 2 y^2) z + (y^4 - 7 y^2 + 6)$.
Now, for the proposed factor really to be a factor, this remainder must be the zero polynomial, that is: \begin{align*} 2 - 2 y^2 &= 0 \\ y^4 - 7 y^2 + 6 &= 0 \end{align*} From the first one you get $y = 1$, and that checks with the second one, so your roots are $1 \pm i$, and the factor is $z^2 - 2 z + 2$. The other one is $z^2 + 7$, with respective roots $\pm i \sqrt{7}$.