$\displaystyle x^2+x+1=0$

2. Originally Posted by Pre-Calculus
$\displaystyle x^2+x+1=0$
$\displaystyle x=\frac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)}$

It ain't looking good...

3. What do you mean?

4. Originally Posted by Pre-Calculus
What do you mean?
I mean that this particular equation has no real zeros.

IE $\displaystyle x^2+x+1\neq0$ for all x.

The discriminant in the quadratic formula conveys this.

5. Originally Posted by Pre-Calculus
$\displaystyle x^2+x+1=0$
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": $\displaystyle a^2- b^2= (a-b)(a+b)$.

$\displaystyle x^2+ x+ 1= (x+ \frac{1}{2}- i\frac{\sqrt{3}}{2})(x+\frac{1}{2}+ i\frac{\sqrt{3}}{2})$