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QLDSpecialist MathematicsSyllabus dot point

Topic 3: Complex numbers

Factorise polynomials with real coefficients over the complex field, apply the conjugate root theorem and the fundamental theorem of algebra, and find all roots of polynomial equations including those with complex coefficients

A focused answer to the QCE Specialist Mathematics Unit 3 dot point on factorising polynomials over the complex numbers. Covers the fundamental theorem of algebra, the conjugate root theorem for real coefficients, dividing out known factors and reconstructing real quadratic factors, with a verified worked example and the conjugate-pairs trap.

Generated by Claude Opus 4.76 min answer

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What this dot point is asking

QCAA wants you to find every root of a polynomial, real or complex, and to factorise it completely over the complex field. The key tools are the fundamental theorem of algebra, which guarantees the right number of roots, and the conjugate root theorem, which says complex roots of a real polynomial come in conjugate pairs. This lets you use one known complex root to unlock the rest.

The answer

Fundamental theorem of algebra

Every polynomial of degree n≥1n \geq 1 with complex coefficients has exactly nn roots in the complex numbers, counted with multiplicity. Consequently a degree-nn polynomial factorises completely into nn linear factors over C\mathbb{C}:

P(z)=a(z−z1)(z−z2)⋯(z−zn).P(z) = a(z - z_1)(z - z_2)\cdots(z - z_n).

Conjugate root theorem

If a polynomial has real coefficients and z=α+iβz = \alpha + i\beta is a root, then its conjugate zˉ=α−iβ\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.

Real quadratic factors from a conjugate pair

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

(z−(α+iβ))(z−(α−iβ))=z2−2αz+(α2+β2).(z - (\alpha + i\beta))(z - (\alpha - i\beta)) = z^2 - 2\alpha z + (\alpha^2 + \beta^2).

So every real polynomial factorises over the reals into real linear factors and real irreducible quadratics, and each such quadratic splits into a conjugate pair over C\mathbb{C}.

Strategy for finding all roots

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.

Sum and product of roots

For P(z)=anzn+⋯+a0P(z) = a_n z^n + \dots + a_0, the sum of the roots is −an−1/an-a_{n-1}/a_n and the product is (−1)na0/an(-1)^n a_0/a_n. These relations check your factorisation: the roots you find must reproduce the coefficients.

Complex coefficients

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 C\mathbb{C}) or by substitution, without assuming conjugate pairs.