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

##### 1 Answer

where

#### Explanation:

Given:

#x^4-x^2-x+1#

First notice that the sum of the coefficients is

#1-1-1+1 = 0#

Hence we can tell that

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

Factoring the remaining cubic is somewhat more complicated:

**Discriminant**

The discriminant

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

In our example,

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

Since

**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

**Cardano's method**

We want to solve:

#t^3-3t-25=0#

Let

Then:

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

Add the constraint

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

Multiply through by

#(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

#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

Now

#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))#