<s>[INST] How do I completely solve the equation $z^4 - 2z^3 + 9z^2 - 14z + 14 = 0$ where there is a root with the real part of $1$. I would please like some help with solving the following equation:
$$z^4 - 2z^3 + 9z^2 - 14z + 14 = 0$$
All I know about the equation is that there is a root with the real part of $1$.
My approach has been to factor out the root: $1 + yi$ and divide the equation with it. The following calculations (using the division algorithm) is quite cumbersome, and I wonder if there is any better way of doing this?
Thank you kindly for your help! 
 [/INST] As the coefficients of the different powers of $z$ are real,
if one of the four roots is $1+yi,$ the other must be its conjugate $1-yi$ .
If the other two roots are $x,w,$  
then using Vieta's Formulas for the coefficient of $z^3$, $1+yi+1-yi+x+w=2\implies w=-x$
So,  $$(z-x)(z+x)\{z-(1+yi)\}\{z-(1-yi)\}=0$$
$$\implies (z^2-x^2)(z^2-2z+1+y^2)=0$$
$$\implies z^4-2z^3+z^2(1+y^2-x^2)+2x^2z-x^2(1+y^2)=0$$
Now compare the coefficients of the different powers of $z$ 
So, $2x^2=-14,x^2=-7,x=\pm \sqrt{-7}=\pm \sqrt7i$ and $1+y^2-x^2=9\implies 1+y^2+7=9,y^2=1$
So, the roots are $\pm \sqrt7i,1\pm i$
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