Question

How do you factor `x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x +1`?

Answer 1

`x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1`

`= (x^2+x+1)(x^2-2cos((2 pi)/9)x+1)(x^2-2cos((4 pi)/9)x+1)(x^2-2cos((8 pi)/9)x+1)`

Explanation:

Notice that if you multiply this polynomial by `(x-1)` the result is `x^9-1`.

So the zeros of this polynomial are all the Complex `9`th roots of `1` except `1`.

Using the difference of cubes identity a couple of times, we find:

`x^9-1 = (x^3)^3-1^3 = (x^3-1)(x^6+x^3+1) = (x-1)(x^2+x+1)(x^6+x^3+1)`

Then we can divide both ends by `(x-1)` to get:

`x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 = (x^2+x+1)(x^6+x^3+1)`

No surprise there.

Note that the zeros of `x^2+x+1` are the Complex `3`rd roots of `1`.

The zeros of `x^6+x^3+1` are the Complex `9`th roots that are not `3`rd roots, i.e.

`e^((2k pi)/9i) = cos((2k pi)/9) + i sin((2k pi)/9)`

with `k = +-1, +-2, +-4`

The factors with Real coefficients, composed of Complex conjugate pairs are:

`(x-cos((2k pi)/9)-i sin((2k pi)/9))(x-cos((2k pi)/9)+i sin((2k pi)/9))`

`=x^2-2cos((2k pi)/9)x+1`

for `k = 1, 2, 4`

Hence:

`x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1`

`= (x^2+x+1)(x^2-2cos((2 pi)/9)x+1)(x^2-2cos((4 pi)/9)x+1)(x^2-2cos((8 pi)/9)x+1)`