How do you find all roots $x^4 - 4x^3 + 6x^2 - 4x + 5$ given one root 2 -i?

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Answer 1 · 2016-08-14T13:23:27

This quartic has zeros $2+-i$ and $+-i$

$f(x) = x^4-4x^3+6x^2-4x+5$

Since the coefficients of $f(x)$ are all Real, any Complex zeros must occur in Complex conjugate pairs.

So if $2-i$ is a zero then so is $2+i$ , and we have corresponding factors $(x-2+i)$ and $(x-2-i)$

Then:

$(x-2+i)(x-2-i) = (x-2)^2-i^2 = x^2-4x+5$

We find:

$x^4-4x^3+6x^2-4x+5 = (x^2-4x+5)(x^2+1)$

The remaining zeros, which are those of $x^2+1$ are simply $+-i$

How do you find all roots x^4 - 4x^3 + 6x^2 - 4x + 5x^4 - 4x^3 + 6x^2 - 4x + 5 given one root 2 -i?