Question

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

Answer 1

`x^2+x+1` has no real factors.

To ascertain this, notice that it is of the form `a^2+bx+c` with `a = b = c = 1`. This has determinant calculated as follows:

`Delta = b^2-4ac = 1^2-(4xx1xx1) = 1-4 = -3`

Since the determinant is negative, the polynomial has no zeros for real values of `x` and therefore no factors with real coefficients.

On the other hand, if you are allowed to use complex coefficients, they can be calculated from the roots of `x^2+x+1=0` as follows:

`x =(-b+-sqrt(Delta))/(2a) = (-1+-sqrt(-3))/2 = (-1+-isqrt(3))/2`

These are the so-called primitive cube roots of unity, often denoted by the symbols `omega` and `omega^2.`

`omega^3 = 1`