Quick questions on Factorising polynomials over the complex field in QCE Specialist Mathematics Unit 3

If a polynomial has real coefficients and $z = \alpha + i\beta$ is a root, then its conjugate $\bar z = \alpha - i\beta$ is also a root. Complex roots of real polynomials therefore occur in conjugate pairs. This is why a real cubic has either three real roots or one real root plus a conjugate pair; complex roots cannot appear alone.

A conjugate pair $\alpha \pm i\beta$ multiplies to a quadratic with real coefficients:

Given one complex root of a real polynomial, immediately write down its conjugate as a second root. Multiply the two corresponding factors to get a real quadratic, divide the polynomial by that quadratic (polynomial long division), and solve the remaining lower-degree quotient. This reduces a quartic to a quadratic in one division.

When coefficients are themselves complex, the conjugate root theorem no longer applies, since it requires real coefficients. You then find roots directly, for example by the quadratic formula (which holds over $\mathbb{C}$) or by substitution, without assuming conjugate pairs.

The coefficients are real, so $\bar z = 1 - 2i$ is also a root.

With $\alpha = 1$, $\beta = 2$:

Polynomial division gives

From $z - 1 = 0$, the third root is $z = 1$. The complete set is $z = 1$, $z = 1 + 2i$, $z = 1 - 2i$.

$1 + (1 + 2i) + (1 - 2i) = 3$, matching $-(-3)/1 = 3$. The factorisation over $\mathbb{C}$ is $(z - 1)(z - 1 - 2i)(z - 1 + 2i)$.