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

Jan 20, 2017

$- 2 {x}^{3} + 6 {x}^{2} - 3 = - 2 \left(x - {x}_{0}\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

where:

${x}_{n} = 1 + 2 \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{1}{4}\right) + \frac{2 n \pi}{3}\right)$

#### Explanation:

Given:

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

Rational roots theorem

By the rational roots 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 $- 3$ and $q$ a divisor of the coefficient $- 2$ of the leading term.

That means that the only possible rational zeros are:

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

Trying each in ascending order, we find:

$f \left(- 3\right) = 105$

$f \left(- \frac{3}{2}\right) = \frac{69}{4}$

$f \left(- 1\right) = 5$

$f \left(- \frac{1}{2}\right) = - \frac{5}{4}$

$f \left(\frac{1}{2}\right) = - \frac{7}{4}$

$f \left(1\right) = 1$

$f \left(\frac{3}{2}\right) = \frac{15}{4}$

$f \left(3\right) = - 3$

So $f \left(x\right)$ has no rational zeros, but it has irrational zeros in each of the intervals: $\left(- 1 , - \frac{1}{2}\right)$, $\left(\frac{1}{2} , 1\right)$, $\left(\frac{3}{2} , 3\right)$

$\textcolor{w h i t e}{}$
Tschirnhaus transformation

To make the cubic easier to solve, use a linear substitution called a Tschirnhaus transformation as follows:

$\left(- 4\right) f \left(x\right) = 8 {x}^{3} - 24 {x}^{2} + 12$

$\textcolor{w h i t e}{\left(- 4\right) f \left(x\right)} = {\left(2 x - 2\right)}^{3} - 12 \left(2 x - 2\right) - 4$

$\textcolor{w h i t e}{\left(- 4\right) f \left(x\right)} = {t}^{3} - 12 t - 4$

where $t = 2 x - 2$

$\textcolor{w h i t e}{}$
Trigonometric substitution

Next we make a substitution:

$t = k \cos \theta$

where $k$ is chosen so that the resulting expansion contains:

$4 {\cos}^{3} \theta - 3 \cos \theta = \cos 3 \theta$

Let $k = 4$

Then:

$0 = {t}^{3} - 12 t - 4$

$\textcolor{w h i t e}{0} = {\left(k \cos \theta\right)}^{3} - 12 \left(k \cos \theta\right) - 4$

$\textcolor{w h i t e}{0} = k \left({k}^{2} {\cos}^{3} \theta - 12 \cos \theta\right) - 4$

$\textcolor{w h i t e}{0} = 4 \left(16 {\cos}^{3} \theta - 12 \cos \theta\right) - 4$

$\textcolor{w h i t e}{0} = 16 \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) - 4$

$\textcolor{w h i t e}{0} = 16 \cos 3 \theta - 4$

Hence:

$\cos 3 \theta = \frac{1}{4}$

So:

$3 \theta = \pm {\cos}^{- 1} \left(\frac{1}{4}\right) + 2 n \pi \text{ }$ for some integer $n$

So:

$\theta = \pm \frac{1}{3} {\cos}^{- 1} \left(\frac{1}{4}\right) + \frac{2 n \pi}{3}$

So:

$\cos \theta = \cos \left(\pm \frac{1}{3} {\cos}^{- 1} \left(\frac{1}{4}\right) + \frac{2 n \pi}{3}\right)$

So:

${t}_{n} = k \cos \theta = 4 \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{1}{4}\right) + \frac{2 n \pi}{3}\right)$

(( we can discard the $\pm$ since $\cos \left(- \alpha\right) = \cos \left(\alpha\right)$ for any $\alpha$ ))

This last formula provides $3$ distinct values ${t}_{n}$, for $n = 0 , 1 , 2$.

Then $x = 1 + \frac{t}{2}$

Hence zeros of our original cubic:

${x}_{n} = 1 + 2 \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{1}{4}\right) + \frac{2 n \pi}{3}\right) \text{ }$ for $n = 0 , 1 , 2$

${x}_{0} \approx 2.8100379$

${x}_{1} \approx - 0.6417835$

${x}_{2} \approx 0.8317456$

Then $f \left(x\right)$ can be factorised as:

$- 2 {x}^{3} + 6 {x}^{2} - 3 = - 2 \left(x - {x}_{0}\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$