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

##### 3 Answers

To find this, first notice that

Without bothering to multiply this out fully just yet, notice that the coefficient of

...

Next, the coefficient of

...

The constant term gives us

...

Note that the coefficient of

Finally notice that the coefficient of

#### Answer:

#### Explanation:

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

What are the zeros of

The real zeros are

Here are all

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 *imaginary unit*, satisfying

For instance we find:

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

That's the zero

If

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#

#### Answer:

#### Explanation:

Given:

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