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

1 Answer
George C.
May 9, 2017

#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)()#

Related questions
See all questions in Factoring Completely
Impact of this question
1140 views around the world
You can reuse this answer
Creative Commons License