1. **State the problem:** We want to factor the quadratic expression $x^2 + x + 1$. 2. **Recall the factoring formula:** For a quadratic $ax^2 + bx + c$, the factors are found by solving $ax^2 + bx + c = 0$ or by finding two numbers that multiply to $ac$ and add to $b$. 3. **Calculate the discriminant:** The discriminant is $\Delta = b^2 - 4ac$. Here, $a=1$, $b=1$, and $c=1$, so $$\Delta = 1^2 - 4 \times 1 \times 1 = 1 - 4 = -3.$$ 4. **Interpret the discriminant:** Since $\Delta < 0$, there are no real roots, so the quadratic cannot be factored over the real numbers into linear factors. 5. **Factor over complex numbers:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}.$$ 6. **Write the factorization:** $$x^2 + x + 1 = \left(x - \frac{-1 + i\sqrt{3}}{2}\right)\left(x - \frac{-1 - i\sqrt{3}}{2}\right) = \left(x - \frac{-1}{2} - \frac{i\sqrt{3}}{2}\right)\left(x - \frac{-1}{2} + \frac{i\sqrt{3}}{2}\right).$$ **Final answer:** The quadratic $x^2 + x + 1$ is not factorable over the real numbers but factors over the complex numbers as $$\left(x + \frac{1}{2} - \frac{i\sqrt{3}}{2}\right)\left(x + \frac{1}{2} + \frac{i\sqrt{3}}{2}\right).$$