How do you factor #x^4+x^2+1#?

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

To find this, first notice that #x^4 + x^2 + 1 > 0# for all (real) values of #x#. So there are no linear factors, only quadratic ones.

#x^4 + x^2 + 1 = (ax^2 + bx + c)(dx^2 + ex + f)#

Without bothering to multiply this out fully just yet, notice that the coefficient of #x^4# gives us #ad = 1#. We might as well let #a = 1# and #d = 1#.

... #= (x^2 + bx + c)(x^2 + ex + f)#

Next, the coefficient of #x^3# gives us #b + e = 0#, so #e = -b#.

... #= (x^2 + bx + c)(x^2 -bx + f)#

The constant term gives us #cf = 1#, so either #c = f = 1# or #c = f = -1#. Let's try #c = f = 1#.

... #= (x^2 + bx + 1)(x^2 - bx + 1)#

Note that the coefficient of #x# will vanish nicely when these are multiplied out.

Finally notice that the coefficient of #x^2# is #(1 - b^2 + 1) = 2 - b^2#, giving us #1 = 2 - b^2#, thus #b^2 = 1#, so #b = 1# or #b = -1#.

This really makes a bit more sense in the complex numbers...

First note that:

#(x^2-1)(x^4+x^2+1) = x^6-1#

So zeros of #x^4+x^2+1# are also zeros of #x^6-1#.

What are the zeros of #x^6-1#?

The real zeros are #1# and #-1#, which are zeros of #x^2-1#, the factor we introduced. So the four zeros of #x^4+x^2+1# are the four complex zeros of #x^6-1# apart from #+-1#.

Here are all #6# in the complex plane:

graph{((x-1)^2+(y-0)^2-0.002)((x-1/2)^2+(y-sqrt(3)/2)^2-0.002)((x+1/2)^2+(y-sqrt(3)/2)^2-0.002)((x+1)^2+(y-0)^2-0.002)((x+1/2)^2+(y+sqrt(3)/2)^2-0.002)((x-1/2)^2+(y+sqrt(3)/2)^2-0.002) = 0 [-2.5, 2.5, -1.25, 1.25]}

They form the vertices of a regular hexagon.

de Moivre's formula tells us that:

#(cos theta + i sin theta)^n = cos n theta + i sin theta#

where #i# is the imaginary unit, satisfying #i^2=-1#

For instance we find:

#(cos (pi/3) + i sin (pi/3))^6 = cos 2pi + i sin 2pi = 1 + 0 = 1#

That's the zero #1/2+sqrt(3)/2i# that we see in Q1.

If #a# is a zero of a polynomial then #(x-a)# is a factor.

Hence:

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

#color(white)(x^4+x^2+1) = (x^2-x+1)(x^2+x+1)#

For example:

#(x-1/2-sqrt(3)/2i)(x-1/2+sqrt(3)/2i)#

#=x^2-((1/2+color(red)(cancel(color(black)(sqrt(3)/2i))))+(1/2-color(red)(cancel(color(black)(sqrt(3)/2i)))))x+(1/2+sqrt(3)/2i)(1/2-sqrt(3/2)i)#

#=x^2-x+((1/2)^2-(sqrt(3)/2i)^2)#

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

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

#=x^2-x+1#

Given: #x^4+x^2+1#

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

The difference of cubes identity can be written:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

The sum of cubes identity can be written:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

Note that:

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

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

#color(white)((x^2-1)(x^4+x^2+1)) = x^6-1#

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

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

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

#color(white)((x^2-1)(x^4+x^2+1)) = (x-1)(x^2+x+1)(x+1)(x^2-x+1)#

#color(white)((x^2-1)(x^4+x^2+1)) = (x^2-1)(x^2+x+1)(x^2-x+1)#

Dividing both ends by #(x^2-1)# we get:

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