Question

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

Answer 1

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

`color(green)(2*x)*x*x+2*color(green)(2*x)*x-2*2*color(green)(2*x)`

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

`2x(x^2+2x-4)`

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:

`2x(x+(1+sqrt(5)))(x+(1-sqrt(5)))`

Answer 2

`2x^3+4x^2-8x = 2x(x^2+2x-4)`

`x^2+2x-4` is of the form `ax^2+bx+c` with `a=1`, `b=2` and `c=-4`.

The discriminant of this quadratic is

`Delta = b^2-4ac = 2^2-4xx1xx-4 = 4+16 = 20`

Since this is positive, the quadratic equation `x^2+2x-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+2x-4 = 0` are:

`x = (-b+-sqrt(Delta))/(2a)`

`=(-2+-sqrt(20))/2`

`=(-2+-2sqrt(5))/2`

`=-1+-sqrt(5)`

So `(x - (-1+sqrt(5))) = (x+1-sqrt(5))`

and `(x - (-1-sqrt(5))) = (x+1+sqrt(5))` are both factors.

`x^2+2x-4 = (x+1-sqrt(5))(x+1+sqrt(5))`

So if we are allowed irrational factors,

`2x^3+4x^2-8x = 2x(x+1-sqrt(5))(x+1+sqrt(5))`

Otherwise, we have to stop at

`2x^3+4x^2-8x = 2x(x^2+2x-4)`