Question

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

Answer 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`.

Answer 2

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

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 `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`

Answer 3

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

Explanation:

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)`