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

Step-by-step solution

1. 1

   Compute the discriminant to settle the real case. With $a=1$, $b=-1$, $c=1$:

   $\Delta=b^{2}-4ac=(-1)^{2}-4(1)(1)=1-4=-3$

   Since $\Delta<0$ the quadratic has no real roots, so it is irreducible over $\mathbb{R}$ — no pair of real linear factors exists, and searching for integer factor pairs of $1$ is futile.

2. 2

   Find the complex roots with the quadratic formula.

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

   The two roots are complex conjugates, as they must be for a polynomial with real coefficients.

3. 3

   Write the factorisation over $\mathbb{C}$. A monic quadratic equals the product of $(x-\text{root})$ over its roots:

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

4. 4

   Verify by multiplying back. The product is $x^{2}-(r_{1}+r_{2})x+r_{1}r_{2}$. The sum of the roots is $\dfrac{1+i\sqrt3}{2}+\dfrac{1-i\sqrt3}{2}=1$ ✓ and the product is $\dfrac{1^{2}-\left(i\sqrt3\right)^{2}}{4}=\dfrac{1+3}{4}=1$ ✓, giving $x^{2}-x+1$ exactly. Substituting either root leaves a residual below $10^{-14}$ ✓.

5. 5

   Note the real alternative and the identity behind it. Completing the square gives the real form $x^{2}-x+1=\left(x-\tfrac12\right)^{2}+\tfrac34$, showing the minimum value $\tfrac34>0$ — an independent proof it never vanishes. The roots are also the primitive sixth roots of unity, reflecting the identity $x^{2}-x+1=\dfrac{x^{3}+1}{x+1}$.