Question

How do you factor `x^4-x^2-x-1` completely ?

Answer 1

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

`color(white)(x^4-x^2-x-1) = (x-1)(x-x_1)(x-x_2)(x-x_3)`

where `x_1, x_2, x_3` are the roots found below...

Explanation:

Given:

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

First notice that the sum of the coefficients is `0`. That is:

`1-1-1+1 = 0`

Hence we can tell that `x=1` is a zero and `(x-1)` a factor:

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

Factoring the remaining cubic is somewhat more complicated:

`color(white)()`
Discriminant

The discriminant `Delta` of a cubic polynomial in the form `ax^3+bx^2+cx+d` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

In our example, `a=1`, `b=1`, `c=0` and `d=-1`, so we find:

`Delta = 0+0+4-27+0 = -23`

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

`color(white)()`
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(x^3+x^2-1)`

`=27x^3+27x^2-27`

`=(3x+1)^3-3(3x+1)-25`

`=t^3-3t-25`

where `t=(3x+1)`

`color(white)()`
Cardano's method

We want to solve:

`t^3-3t-25=0`

Let `t=u+v`.

Then:

`u^3+v^3+3(uv-1)(u+v)-25=0`

Add the constraint `v=1/u` to eliminate the `(u+v)` term and get:

`u^3+1/u^3-25=0`

Multiply through by `u^3` and rearrange slightly to get:

`(u^3)^2-25(u^3)+1=0`

Use the quadratic formula to find:

`u^3=(25+-sqrt((-25)^2-4(1)(1)))/(2*1)`

`=(25+-sqrt(625-4))/2`

`=(25+-sqrt(621))/2`

`=(25+-3sqrt(69))/2`

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=root(3)((25+3sqrt(69))/2)+root(3)((25-3sqrt(69))/2)`

and related Complex roots:

`t_2=omega root(3)((25+3sqrt(69))/2)+omega^2 root(3)((25-3sqrt(69))/2)`

`t_3=omega^2 root(3)((25+3sqrt(69))/2)+omega root(3)((25-3sqrt(69))/2)`

where `omega=-1/2+sqrt(3)/2i` is the primitive Complex cube root of `1`.

Now `x=1/3(-1+t)`. So the roots of our original cubic are:

`x_1 = 1/3(-1+root(3)((25+3sqrt(69))/2)+root(3)((25-3sqrt(69))/2))`

`x_2 = 1/3(-1+omega root(3)((25+3sqrt(69))/2)+omega^2 root(3)((25-3sqrt(69))/2))`

`x_3 = 1/3(-1+omega^2 root(3)((25+3sqrt(69))/2)+omega root(3)((25-3sqrt(69))/2))`

`color(white)()`