$\displaystyle x^2+x+1=0$
Normally, "factoring" means "factor into terms having integer coefficients" and, as VonNemo19 said, that cannot be done here.
But if you really want to factor- and aren't concerned about the factors having integer, or even real, coefficients, you can "complete the square":
$\displaystyle x^2+ x+ 1= x^2+ x+ \frac{1}{4}- \frac{1}{4}+ 1= 0$
(Divide the coefficient of x (1) by to (1/2) and square (1/4). Add and subtract that.)
$\displaystyle x^2+ x+ 1= (x+ \frac{1}{2})^2+ \frac{3}{4}$
and think of this as a "difference of squares":
$\displaystyle (x+ \frac{1}{2})^2- (-\frac{3}{4})= (x+\frac{1}{2})^2- (i \frac{\sqrt{3}}{2})^2$
Which can be factored as "sum and difference": $\displaystyle a^2- b^2= (a-b)(a+b)$.
$\displaystyle x^2+ x+ 1= (x+ \frac{1}{2}- i\frac{\sqrt{3}}{2})(x+\frac{1}{2}+ i\frac{\sqrt{3}}{2})$