Factor x^2 + x + 1 over the reals if possible, otherwise over the complex numbers.

Step-by-step solution

1. 1

   Rule out a real factorisation with the discriminant. With $a=b=c=1$:

   $\Delta=b^{2}-4ac=1-4=-3<0$

   No real roots exist, so $x^{2}+x+1$ is irreducible over $\mathbb{R}$. Completing the square confirms it constructively: $x^{2}+x+1=\left(x+\tfrac12\right)^{2}+\tfrac34\ge\tfrac34>0$, so the expression never reaches zero.

2. 2

   Find the complex roots.

   $x=\frac{-1\pm\sqrt{-3}}{2}=\frac{-1\pm i\sqrt{3}}{2}$

   a conjugate pair, as always for real coefficients.

3. 3

   Write the factorisation over $\mathbb{C}$.

   $x^{2}+x+1=\left(x-\frac{-1+i\sqrt{3}}{2}\right)\left(x-\frac{-1-i\sqrt{3}}{2}\right)=\left(x+\frac12-\frac{i\sqrt3}{2}\right)\left(x+\frac12+\frac{i\sqrt3}{2}\right)$

4. 4

   Verify with Vieta. The roots sum to $\dfrac{-1+i\sqrt3}{2}+\dfrac{-1-i\sqrt3}{2}=-1=-\dfrac{b}{a}$ ✓ and multiply to $\dfrac{(-1)^{2}-\left(i\sqrt3\right)^{2}}{4}=\dfrac{1+3}{4}=1=\dfrac{c}{a}$ ✓, so expanding the product returns $x^{2}+x+1$ exactly. Substituting either root gives a residual below $10^{-15}$ ✓.

5. 5

   Recognise the roots as cube roots of unity. Because

   $x^{3}-1=(x-1)\left(x^{2}+x+1\right)\qquad\Longrightarrow\qquad x^{2}+x+1=\frac{x^{3}-1}{x-1}$

   the two roots are exactly the cube roots of $1$ other than $1$ itself — written $\omega$ and $\omega^{2}$. A numerical check confirms $\omega^{3}=1$ and $\omega^{2}+\omega+1=0$ to within $10^{-15}$ ✓. This identity is also why $x^{2}+x+1$ shows up whenever a difference of cubes is factored.