Question

What is the factored form of `2x^3 + 4x^2 – x?`

Answer 1

`color(blue)((2x)(x+(2-sqrt(6))/(2))(x+(2+sqrt(6))/(2))`

Explanation:

`2x^3+4x^2-x`

First factor out `x`:

`x(2x^2+4x-1)`

Looking at the factor:

`2x^2+4x-1`

It is not possible to factor this using the straight forward method. We will have to find the roots to this and work backwards.

First we recognise the if `alpha` and `beta` are the two roots, then:

`a(x-alpha)(x-beta)` are factors of `2x^2+4x-1`

Where `a` is a multiplier:

Roots of `2x^2+4x-1=0` using quadratic formula:

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

`x=(-4+-sqrt(24))/(4)`

`x=(-4+-2sqrt(6))/(4)=x=(-2+-sqrt(6))/(2)`

`x=(-2+sqrt(6))/(2)`

`x=(-2-sqrt(6))/(2)`

So we have:

`a(x-((-2+sqrt(6))/(2)))(x-((-2-sqrt(6))/(2)))`

`a(x+(2-sqrt(6))/(2))(x+(2+sqrt(6))/(2))`

We can see by the coefficient of `x^2` in `2x^2+4x-1` that:

`a=2`

`:.`

`2(x+(2-sqrt(6))/(2))(x+(2+sqrt(6))/(2))`

And including the factor `x` from earlier:

`(2x)(x+(2-sqrt(6))/(2))(x+(2+sqrt(6))/(2))`

I am not sure if this is what you were looking for. This method isn't particularly useful, since often the point of factoring is to find the roots and here we have to find the roots to find the factors. Factoring higher order polynomials can be difficult if the factors aren't rational as in this case.