How do you factor 3x^3 - 10x^2 + 2 = 0?

May 6, 2016

$3 {x}^{3} - 10 {x}^{2} + 2 = 3 \left(x - {x}_{0}\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

where:

${x}_{k} = \frac{\sqrt{15}}{10} \sec \left(\frac{2 k \pi}{3} + \frac{1}{3} \arccos \left(- \frac{9 \sqrt{15}}{100}\right)\right)$ for $k = 0 , 1 , 2$

$\left\{\begin{matrix}{x}_{0} \approx 0.48368 \\ {x}_{1} \approx - 0.42137 \\ {x}_{2} \approx 3.271026\end{matrix}\right.$

Explanation:

Given:

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

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p$, $q$ with $p$ a divisor of the constant term $2$ and $q$ a divisor of the coefficient $3$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{3}$, $\pm \frac{2}{3}$, $\pm 1$, $\pm 2$

None of these are zeros of $f \left(x\right)$, so it has no rational zeros.

Note that:

$\left\{\begin{matrix}f \left(- 1\right) = - 3 - 10 + 2 = - 11 < 0 \\ f \left(0\right) = 0 + 0 + 2 = 2 > 0 \\ f \left(1\right) = 3 - 10 + 2 = - 5 < 0 \\ f \left(4\right) = 192 - 160 + 2 = 34 > 0\end{matrix}\right.$

So $f \left(x\right)$ has three Real zeros.

I will use a trigonometric method to derive expressions for the three zeros:

Let $t = \frac{1}{x}$.

Then the zeros of $f \left(x\right)$ are the roots of:

$2 {t}^{3} - 10 t + 3 = 0$

Let $t = \frac{2 \sqrt{15}}{3} \cos \theta$

The multiplier here is chosen to result in a match for the formula for $\cos 3 \theta$...

Then:

$0 = 2 {t}^{3} - 10 t + 3$

$= 2 {\left(\frac{2 \sqrt{15}}{3} \cos \theta\right)}^{3} - 10 \left(\frac{2 \sqrt{15}}{3} \cos \theta\right) + 3$

$= \frac{20 \sqrt{15}}{9} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) + 3$

$= \frac{20 \sqrt{15}}{9} \cos 3 \theta + 3$

So:

$\cos 3 \theta = - \frac{27}{20 \sqrt{15}} = - \frac{9 \sqrt{15}}{100}$

So

$3 \theta = 2 k \pi \pm \arccos \left(- \frac{9 \sqrt{15}}{100}\right)$

So:

$t = \frac{2 \sqrt{15}}{3} \cos \theta = \frac{2 \sqrt{15}}{3} \cos \left(\frac{2 k \pi}{3} \pm \frac{1}{3} \arccos \left(- \frac{9 \sqrt{15}}{100}\right)\right)$

This gives us three distinct roots:

$t = \frac{2 \sqrt{15}}{3} \cos \left(\frac{2 k \pi}{3} + \frac{1}{3} \arccos \left(- \frac{9 \sqrt{15}}{100}\right)\right)$ for $k = 0 , 1 , 2$

Then:

$x = \frac{1}{t} = \frac{3}{2 \sqrt{15}} \sec \left(\frac{2 k \pi}{3} + \frac{1}{3} \arccos \left(- \frac{9 \sqrt{15}}{100}\right)\right)$ for $k = 0 , 1 , 2$

$= \frac{\sqrt{15}}{10} \sec \left(\frac{2 k \pi}{3} + \frac{1}{3} \arccos \left(- \frac{9 \sqrt{15}}{100}\right)\right)$ for $k = 0 , 1 , 2$

Call these three roots ${x}_{0}$, ${x}_{1}$ and ${x}_{2}$.

Then $f \left(x\right) = 3 \left(x - {x}_{0}\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

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Footnote

Why did I use a trigonometric method to solve this cubic?

When a cubic has $3$ Real roots, methods like Cardano's result in irreducible cube roots of Complex numbers. This is called "casus irreducibilis". The result of such cube roots cannot be expressed in the form $a + b i$ using square and/or cube roots of Real numbers.

Since we are trying to describe Real roots, expressions involving explicit Complex arithmetic seem somewhat out of place.

In addition, note that taking cube roots relates in an essential way to trisecting angles, so trigonometry is appropriate. The trigonometric solution brings this out.