Question

How do you factor `(x^6)+1`?

Answer 1

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

How did I find this? First of all, substitute `x^2 = y` to get `y^3 + 1`.

Notice that `y = -1` is a solution of `y^3+1 = 0`, so `(y+1)` is a factor.

We can easily derive `y^3+1=(y^2-y+1)(y+1)`, noticing the way that the intermediate terms cancel out when you multiply this out.

Substituting `y = x^2` back in, we get:
`x^6+1=(x^4-x^2+1)(x^2+1)`

How do we know this factorization is complete?

There is no linear factor, because `x^6+1` is always > 0 (for real values of `x`).

How about quadratic factors?

Suppose `x^4-x^2+1=(x^2+ax+b)(x^2+cx+d)`.

Looking at the coefficient of `x^3`, we get `a+c = 0`, so `c = -a`:

... `=(x^2+ax+b)(x^2-ax+d)`

The constant term gives us that `bd = 1` so `b = d = 1` or `b = d = -1`.
The coefficient of `x^2` gives us `a^2-(b+d)=1`. Hence `a^2 = 3` or `a^2 = -1`. Neither have integer solutions.