# How do you factor completely f(x) = x^5-2 i x^4-(5+3 i) x^3-(7-3 i) x^2+(6+11 i) x-(1+3 i) ?

Dec 14, 2016

$f \left(x\right) = \left(x - 1 - i\right) \left(x - 2 - i\right) \left(x + 1 - \sqrt[3]{2}\right) \left(x + 1 - \omega \sqrt[3]{2}\right) \left(x + 1 - {\omega}^{2} \sqrt[3]{2}\right)$

#### Explanation:

Given:

$f \left(x\right) = {x}^{5} - 2 i {x}^{4} - \left(5 + 3 i\right) {x}^{3} - \left(7 - 3 i\right) {x}^{2} + \left(6 + 11 i\right) x - \left(1 + 3 i\right)$

Use a Gaussian integer version of the rational roots theorem.

Any Gaussian rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for Gaussian integers $p , q$ with $p$ a divisor of the constant term $- \left(1 + 3 i\right)$ and $q$ a divisor of the coefficient $1$ of the leading term.

Note that:

$\left\mid - \left(1 + 3 i\right) \right\mid = \sqrt{{1}^{2} + {3}^{2}} = \sqrt{10} = \sqrt{2} \sqrt{5}$

So $1 + 3 i$ is divisible by some Gaussian integers of modulus $\sqrt{2}$ and $\sqrt{5}$

$\frac{1 + 3 i}{1 + i} = \frac{\left(1 + 3 i\right) \left(1 - i\right)}{\left(1 + i\right) \left(1 - i\right)} = \frac{4 + 2 i}{2} = 2 + i$

Hence the possible Gaussian rational zeros of $f \left(x\right)$ are:

$\pm 1$, $\pm i$, $\pm \left(1 + i\right)$, $\pm \left(1 - i\right)$, $\pm \left(2 + i\right)$, $\pm \left(1 - 2 i\right)$, $\pm \left(1 + 3 i\right)$, $\pm \left(3 - i\right)$

Trying each in turn, we find:

$f \left(1\right) = 1 - 2 i - \left(5 + 3 i\right) - \left(7 - 3 i\right) + \left(6 + 11 i\right) - \left(1 + 3 i\right) = - 6 + 6 i$

$f \left(- 1\right) = - 1 - 2 i + \left(5 + 3 i\right) - \left(7 - 3 i\right) - \left(6 + 11 i\right) - \left(1 + 3 i\right) = - 10 - 10 i$

$f \left(i\right) = {\left(i\right)}^{5} - 2 i {\left(i\right)}^{4} - \left(5 + 3 i\right) {\left(i\right)}^{3} - \left(7 - 3 i\right) {\left(i\right)}^{2} + \left(6 + 11 i\right) \left(i\right) - \left(1 + 3 i\right) = - 8 + 4 i$

$f \left(- i\right) = {\left(- i\right)}^{5} - 2 i {\left(- i\right)}^{4} - \left(5 + 3 i\right) {\left(- i\right)}^{3} - \left(7 - 3 i\right) {\left(- i\right)}^{2} + \left(6 + 11 i\right) \left(- i\right) - \left(1 + 3 i\right) = 20 - 20 i$

$f \left(1 + i\right) = {\left(1 + i\right)}^{5} - 2 i {\left(1 + i\right)}^{4} - \left(5 + 3 i\right) {\left(1 + i\right)}^{3} - \left(7 - 3 i\right) {\left(1 + i\right)}^{2} + \left(6 + 11 i\right) \left(1 + i\right) - \left(1 + 3 i\right)$

$\textcolor{w h i t e}{f \left(1 + i\right)} = \left(1 + 5 i - 10 - 10 i + 5 + i\right) - 2 i {\left(1 + 4 i - 6 - 4 i + 1\right)}^{4} - \left(5 + 3 i\right) {\left(1 + 3 i - 3 - i\right)}^{3} - \left(7 - 3 i\right) {\left(1 + 2 i - 1\right)}^{2} + \left(6 + 11 i\right) \left(1 + i\right) - \left(1 + 3 i\right)$

$\textcolor{w h i t e}{f \left(1 + i\right)} = \left(- 4 - 4 i\right) - 2 i \left(- 4\right) - \left(5 + 3 i\right) \left(- 2 + 2 i\right) - \left(7 - 3 i\right) \left(2 i\right) + \left(6 + 11 i\right) \left(1 + i\right) - \left(1 + 3 i\right)$

$\textcolor{w h i t e}{f \left(1 + i\right)} = 0$

So $x = 1 + i$ is a zero and $\left(x - 1 - i\right)$ a factor:

${x}^{5} - 2 i {x}^{4} - \left(5 + 3 i\right) {x}^{3} - \left(7 - 3 i\right) {x}^{2} + \left(6 + 11 i\right) x - \left(1 + 3 i\right)$

$= \left(x - 1 - i\right) \left({x}^{4} + \left(1 - i\right) {x}^{3} - \left(3 + 3 i\right) {x}^{2} - \left(7 + 3 i\right) x + \left(2 + i\right)\right)$

Let $g \left(x\right) = {x}^{4} + \left(1 - i\right) {x}^{3} - \left(3 + 3 i\right) {x}^{2} - \left(7 + 3 i\right) x + \left(2 + i\right)$

To cut a long story a little shorter, we find:

$g \left(2 + i\right) = {\left(2 + i\right)}^{4} + \left(1 - i\right) {\left(2 + i\right)}^{3} - \left(3 + 3 i\right) {\left(2 + i\right)}^{2} - \left(7 + 3 i\right) \left(2 + i\right) + \left(2 + i\right) = 0$

and

${x}^{4} + \left(1 - i\right) {x}^{3} - \left(3 + 3 i\right) {x}^{2} - \left(7 + 3 i\right) x + \left(2 + i\right)$

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

Then:

${x}^{3} + 3 {x}^{2} + 3 x - 1 = {x}^{3} + 3 {x}^{2} + 3 x + 1 - 2$

$\textcolor{w h i t e}{{x}^{3} + 3 {x}^{2} + 3 x - 1} = {\left(x + 1\right)}^{3} - {\left(\sqrt[3]{2}\right)}^{3}$

$\textcolor{w h i t e}{{x}^{3} + 3 {x}^{2} + 3 x - 1} = \left(\left(x + 1\right) - \sqrt[3]{2}\right) \left(\left(x + 1\right) - \omega \sqrt[3]{2}\right) \left(\left(x + 1\right) - {\omega}^{2} \sqrt[3]{2}\right)$

$\textcolor{w h i t e}{{x}^{3} + 3 {x}^{2} + 3 x - 1} = \left(x + 1 - \sqrt[3]{2}\right) \left(x + 1 - \omega \sqrt[3]{2}\right) \left(x + 1 - {\omega}^{2} \sqrt[3]{2}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.