# How do you factor 2x^3+2x^2-8x+8?

Aug 18, 2016

$2 {x}^{3} + 2 {x}^{2} - 8 x + 8 = 2 \left({x}^{3} + {x}^{2} - 4 x + 4\right) = 2 \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$

where ${x}_{1} , {x}_{2}$ and ${x}_{3}$ are found below...

#### Explanation:

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

Note that if either of the last two signs were inverted then this cubic would factor by grouping. As it is, it is somewhat more complicated...

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

$\textcolor{w h i t e}{}$
Descriminant

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, $a = 1$, $b = 1$, $c = - 4$ and $d = 4$, so we find:

$\Delta = 16 + 256 - 16 - 432 - 288 = - 464$

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

$\textcolor{w h i t e}{}$
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.

$0 = 27 f \left(x\right) = 27 {x}^{3} + 27 {x}^{2} - 108 x + 108$

$= {\left(3 x + 1\right)}^{3} - 39 \left(3 x + 1\right) + 146$

$= {t}^{3} - 39 t + 146$

where $t = \left(3 x + 1\right)$

$\textcolor{w h i t e}{}$
Cardano's method

We want to solve:

${t}^{3} - 39 t + 146 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 13\right) \left(u + v\right) + 146 = 0$

Add the constraint $v = \frac{13}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} + \frac{2197}{u} ^ 3 + 146 = 0$

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

${\left({u}^{3}\right)}^{2} + 146 \left({u}^{3}\right) + 2197 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{- 146 \pm \sqrt{{\left(146\right)}^{2} - 4 \left(1\right) \left(2197\right)}}{2 \cdot 1}$

$= \frac{146 \pm \sqrt{21316 - 8788}}{2}$

$= \frac{146 \pm \sqrt{12528}}{2}$

$= \frac{146 \pm 12 \sqrt{87}}{2}$

$= 73 \pm 6 \sqrt{87}$

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:

${t}_{1} = \sqrt{73 + 6 \sqrt{87}} + \sqrt{73 - 6 \sqrt{87}}$

and related Complex roots:

${t}_{2} = \omega \sqrt{73 + 6 \sqrt{87}} + {\omega}^{2} \sqrt{73 - 6 \sqrt{87}}$

${t}_{3} = {\omega}^{2} \sqrt{73 + 6 \sqrt{87}} + \omega \sqrt{73 - 6 \sqrt{87}}$

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

Now $x = \frac{1}{3} \left(- 1 + t\right)$. So the zeros of our original cubic are:

${x}_{1} = \frac{1}{3} \left(- 1 + \sqrt{73 + 6 \sqrt{87}} + \sqrt{73 - 6 \sqrt{87}}\right)$

${x}_{2} = \frac{1}{3} \left(- 1 + \omega \sqrt{73 + 6 \sqrt{87}} + {\omega}^{2} \sqrt{73 - 6 \sqrt{87}}\right)$

${x}_{3} = \frac{1}{3} \left(- 1 + {\omega}^{2} \sqrt{73 + 6 \sqrt{87}} + \omega \sqrt{73 - 6 \sqrt{87}}\right)$