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

Aug 14, 2016

This quartic has zeros $2 \pm i$ and $\pm i$

#### Explanation:

$f \left(x\right) = {x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 4 x + 5$

Since the coefficients of $f \left(x\right)$ 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 $\left(x - 2 + i\right)$ and $\left(x - 2 - i\right)$

Then:

$\left(x - 2 + i\right) \left(x - 2 - i\right) = {\left(x - 2\right)}^{2} - {i}^{2} = {x}^{2} - 4 x + 5$

We find:

${x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 4 x + 5 = \left({x}^{2} - 4 x + 5\right) \left({x}^{2} + 1\right)$

The remaining zeros, which are those of ${x}^{2} + 1$ are simply $\pm i$