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Working on a task where I´m supposed to find the roots of $P(z) = z^4 - 6z^3 + 11z^2 - 2z - 10$.

One root is given, $z_0 = 2 - i$.

First, it´s clear another root is $z_1 = 2 + i$.

I know I´m supposed to calculate another polynomial by using $(z - z_0)(z - z_1)$.

The solution tells me that equals $z^{2} - 4z + 5$. But I can´t see how that´s possible? Could someone please explain how this is calculated?

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1 Answer

up vote 4 down vote accepted

Let $z_i,0\le i<4$ are the $4$ solutions.

So using Vieta's Formulae,

$z_0z_1z_2z_3=-10$ but $z_0z_1=(2+i)(2-i)=5$ so, $z_2z_3=-2$

and $z_0+z_1+z_2+z_3=6$ but $z_0+z_1=(2+i)+(2-i)=4$ so, $z_2+z_3=2$

So, $z_2,z_3$ are the roots of $t^2-2t-2=0$


Alternatively, the quadratic equation with roots $2\pm i$ is $\{z-(2+i)\}\{z-(2+i)\}=0$ or $z^2-4z-5=0$

So, $z^4-6z^3+11z^2-2z-10=(z^2-4z-5)(z-z_2)(z-z_2)\implies (z-z_2)(z-z_3)=\frac{z^4-6z^3+11z^2-2z-10}{z^2-4z-5}=z^2-2z-2$

So, $z_2,z_3$ are $1\pm\sqrt 3$

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Thank you, the alternative solution is what I was looking for! –  marsrover Dec 8 '12 at 16:01
    
@marsrover, my pleasure. But what about the 1st method? –  lab bhattacharjee Dec 8 '12 at 16:05
    
Hmm, how did you figure out that the last two is 1±√‾‾3? –  marsrover Dec 8 '12 at 16:15
    
@marsrover, mathworld.wolfram.com/QuadraticFormula.html –  lab bhattacharjee Dec 8 '12 at 16:17
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