# How do you factor 2x^3 + 4x^2 - 8x?

May 16, 2015

First, we must identify the common elements in this function. Let's just disaggregate them for a while:

$\textcolor{g r e e n}{2 \cdot x} \cdot x \cdot x + 2 \cdot \textcolor{g r e e n}{2 \cdot x} \cdot x - 2 \cdot 2 \cdot \textcolor{g r e e n}{2 \cdot x}$

Note that the three of them are multiplied by $2 x$. Therefore, this element comes out:

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

As inside the parenthesis we have a quadratic, we can solve it and find its roots, which are ${x}_{1} = - 1 - \sqrt{5}$ and ${x}_{2} = - 1 + \sqrt{5}$.

We can rewrite these roots as

${x}_{1} + 1 + \sqrt{5} = 0$ and ${x}_{2} + 1 - \sqrt{5} = 0$

...and use these as our factors:

$2 x \left(x + \left(1 + \sqrt{5}\right)\right) \left(x + \left(1 - \sqrt{5}\right)\right)$

May 16, 2015

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

${x}^{2} + 2 x - 4$ is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 2$ and $c = - 4$.

The discriminant of this quadratic is

$\Delta = {b}^{2} - 4 a c = {2}^{2} - 4 \times 1 \times - 4 = 4 + 16 = 20$

Since this is positive, the quadratic equation ${x}^{2} + 2 x - 4 = 0$ has 2 distinct real roots. Unfortunately, $20$ is not a perfect square, so those roots are not rational.

The roots of ${x}^{2} + 2 x - 4 = 0$ are:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{- 2 \pm \sqrt{20}}{2}$

$= \frac{- 2 \pm 2 \sqrt{5}}{2}$

$= - 1 \pm \sqrt{5}$

So $\left(x - \left(- 1 + \sqrt{5}\right)\right) = \left(x + 1 - \sqrt{5}\right)$

and $\left(x - \left(- 1 - \sqrt{5}\right)\right) = \left(x + 1 + \sqrt{5}\right)$ are both factors.

${x}^{2} + 2 x - 4 = \left(x + 1 - \sqrt{5}\right) \left(x + 1 + \sqrt{5}\right)$

So if we are allowed irrational factors,

$2 {x}^{3} + 4 {x}^{2} - 8 x = 2 x \left(x + 1 - \sqrt{5}\right) \left(x + 1 + \sqrt{5}\right)$

Otherwise, we have to stop at

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