# How do you factor 5x^3 -3x^2 +16x-6?

Jan 25, 2017

Use Cardano's method to find:

$5 {x}^{3} - 3 {x}^{2} + 16 x - 6 = 5 \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$

where:

${x}_{n} = \frac{1}{15} \left(3 + {\omega}^{n - 1} \sqrt[3]{972 + 15 \sqrt{58983}} + {\omega}^{1 - n} \sqrt[3]{972 - 15 \sqrt{58983}}\right)$

#### Explanation:

Given:

$f \left(x\right) = 5 {x}^{3} - 3 {x}^{2} + 16 x - 6$

We can find the zeros ${x}_{1}$, ${x}_{2}$ and ${x}_{3}$

Then:

$f \left(x\right) = 5 \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$

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Discriminant

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

In our example, $a = 5$, $b = - 3$, $c = 16$ and $d = - 6$, so we find:

$\Delta = 2304 - 81920 - 648 - 24300 + 25920 = - 78644$

Since $\Delta < 0$ this cubic has $1$ Real zero and $2$ non-Real Complex zeros, which are Complex conjugates of one another.

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Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

$0 = 25 f \left(x\right) = 125 {x}^{3} - 75 {x}^{2} + 400 x - 150$

$= {\left(5 x - 1\right)}^{3} + 77 \left(5 x - 1\right) - 72$

$= {t}^{3} + 77 t - 72$

where $t = \left(5 x - 1\right)$

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Cardano's method

We want to solve:

${t}^{3} + 77 t - 72 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + \left(3 u v + 77\right) \left(u + v\right) - 72 = 0$

Add the constraint $v = - \frac{77}{3 u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} - \frac{456533}{27 {u}^{3}} - 72 = 0$

Multiply through by $27 {u}^{3}$ and rearrange slightly to get:

$27 {\left({u}^{3}\right)}^{2} - 1944 \left({u}^{3}\right) - 456533 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{1944 \pm \sqrt{{\left(- 1944\right)}^{2} - 4 \left(27\right) \left(- 456533\right)}}{2 \cdot 27}$

$= \frac{1944 \pm \sqrt{3779136 + 49305564}}{54}$

$= \frac{1944 \pm \sqrt{53084700}}{54}$

$= \frac{1944 \pm \sqrt{{30}^{2} \cdot 58983}}{54}$

$= \frac{972 \pm 15 \sqrt{58983}}{3} ^ 3$

Since this is Real and the derivation is symmetric in $u$ and $v$, we can use one of these roots for ${u}^{3}$ and the other for ${v}^{3}$ to find Real root:

${t}_{1} = \frac{1}{3} \left(\sqrt[3]{972 + 15 \sqrt{58983}} + \sqrt[3]{972 - 15 \sqrt{58983}}\right)$

and related Complex roots:

${t}_{2} = \frac{1}{3} \left(\omega \sqrt[3]{972 + 15 \sqrt{58983}} + {\omega}^{2} \sqrt[3]{972 - 15 \sqrt{58983}}\right)$

${t}_{3} = \frac{1}{3} \left({\omega}^{2} \sqrt[3]{972 + 15 \sqrt{58983}} + \omega \sqrt[3]{972 - 15 \sqrt{58983}}\right)$

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

Now $x = \frac{1}{5} \left(1 + t\right)$. So the roots of our original cubic are:

${x}_{n} = \frac{1}{15} \left(3 + {\omega}^{n - 1} \sqrt[3]{972 + 15 \sqrt{58983}} + {\omega}^{1 - n} \sqrt[3]{972 - 15 \sqrt{58983}}\right)$