# How do you factor x^3 - 6x^2 + 3x - 10?

May 9, 2016

See explanation...

#### Explanation:

Use Cardano's method.

First a simple Tschirnhaus transformation, to eliminate the term of degree $2$...

$0 = {x}^{3} - 6 {x}^{2} + 3 x - 10 = {\left(x - 2\right)}^{3} - 9 \left(x - 2\right) - 20$

Let $t = x - 2$ to get the simplified cubic equation:

${t}^{3} - 9 t - 20 = 0$

Next substitute $t = u + v$ to get:

${u}^{3} + {v}^{3} + 3 \left(u v - 3\right) \left(u + v\right) - 20 = 0$

Add the constraint $v = \frac{3}{u}$ to eliminate the term in $\left(u + v\right)$ and get:

${u}^{3} + {\left(\frac{3}{u}\right)}^{3} - 20 = 0$

Multiply through by ${u}^{3}$ to get:

${\left({u}^{3}\right)}^{2} - 20 \left({u}^{3}\right) + 27 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{20 \pm \sqrt{{20}^{2} - \left(4 \cdot 1 \cdot 27\right)}}{2 \cdot 1}$

$= 10 \pm \frac{\sqrt{400 - 108}}{2}$

$= 10 \pm \frac{\sqrt{292}}{2}$

$= 10 \pm \frac{\sqrt{4 \cdot 73}}{2}$

$= 10 \pm \sqrt{73}$

The derivation was symmetric in $u$ and $v$, hence (noting $x = t + 2$) we can deduce that the Real zero of the original cubic is:

${x}_{1} = 2 + \sqrt[3]{10 - \sqrt{73}} + \sqrt[3]{10 + \sqrt{73}}$

and Complex zeros:

${x}_{2} = 2 + \omega \sqrt[3]{10 - \sqrt{73}} + {\omega}^{2} \sqrt[3]{10 + \sqrt{73}}$

${x}_{3} = 2 + {\omega}^{2} \sqrt[3]{10 - \sqrt{73}} + \omega \sqrt[3]{10 + \sqrt{73}}$

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

Then:

${x}^{3} - 6 {x}^{2} + 3 x - 10 = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$