What is the conjugate root theorem?

The reason is the conjugate arithmetic from earlier work: taking the conjugate of the whole equation P(z) = 0 replaces each z by z but leaves the real coefficients unchanged, so P( z) = 0 as well. A practical consequence is that real polynomials of odd degree must have at least one real root, because complex roots are used up in pairs.

What are real quadratic factors?

A conjugate pair of roots a bi corresponds to the real quadratic factor

TASSpecialist MathematicsSyllabus dot point

How does allowing complex roots let us factorise every polynomial completely?

Factorise polynomials over the complex field using the conjugate root theorem and the fundamental theorem of algebra.

The fundamental theorem of algebra, conjugate root theorem, and factorising real and complex polynomials into linear and quadratic factors, for TCE Mathematics Specialised Unit 3.

Generated by Claude Opus 4.79 min answerUpdated 2026-05-30

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What this dot point is asking

This dot point is about completing the picture of factorisation. In earlier study you factorised over the real numbers and sometimes got stuck with an irreducible quadratic such as $x^2 + 1$ . Once complex numbers are allowed, that obstacle disappears entirely. The two governing results are the fundamental theorem of algebra and the conjugate root theorem.

So a cubic has three complex roots, a quartic has four, and so on. There are never any missing or unreachable roots once you work in the complex numbers.

The conjugate root theorem

The reason is the conjugate arithmetic from earlier work: taking the conjugate of the whole equation $P(z) = 0$ replaces each $z$ by $\bar z$ but leaves the real coefficients unchanged, so $P(\bar z) = 0$ as well. A practical consequence is that real polynomials of odd degree must have at least one real root, because complex roots are used up in pairs.

Real quadratic factors

A conjugate pair of roots $a \pm bi$ corresponds to the real quadratic factor

\big(z - (a + bi)\big)\big(z - (a - bi)\big) = z^2 - 2az + (a^2 + b^2).

This is why a real polynomial factorises into real linear factors (from real roots) and real irreducible quadratic factors (from conjugate pairs). Recognising this lets you reconstruct a polynomial from partial root information.

Building a polynomial from one complex root

Worked example

A cubic $P(x)$ has real coefficients, leading coefficient $1$ , and one root is $x = 2 + i$ . The third root is $x = 5$ . Find $P(x)$ in expanded form.

By the conjugate root theorem, $2 - i$ is also a root. The three roots are $2 + i$ , $2 - i$ and $5$ .

The conjugate pair gives the real quadratic factor

\big(x - (2 + i)\big)\big(x - (2 - i)\big) = x^2 - 4x + (4 + 1) = x^2 - 4x + 5.

Multiply by the remaining factor $(x - 5)$ :

P(x) = (x - 5)(x^2 - 4x + 5) = x^3 - 4x^2 + 5x - 5x^2 + 20x - 25.

Collect terms:

P(x) = x^3 - 9x^2 + 25x - 25.

A quick check: the sum of the roots is $(2 + i) + (2 - i) + 5 = 9$ , which matches the coefficient $-(-9)$ of $x^2$ .

Factorising a given polynomial

When you are handed a polynomial and one complex root, divide out the corresponding real quadratic, then factor the remaining quotient.

Factorising a quartic over the complex numbers

Worked example

Factorise $z^4 + 1$ completely over the complex numbers.

The roots solve $z^4 = -1 = \operatorname{cis}\pi$ . By De Moivre,

z_k = \operatorname{cis}\!\left(\frac{\pi + 2\pi k}{4}\right), \qquad k = 0, 1, 2, 3,

giving arguments $\tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4}$ . These four points are $\tfrac{\sqrt2}{2}(\pm1 \pm i)$ . The complete complex factorisation is the product of $(z - z_k)$ . Pairing conjugate roots gives the real factorisation

z^4 + 1 = (z^2 - \sqrt2\, z + 1)(z^2 + \sqrt2\, z + 1),

since each conjugate pair has sum of real parts $\pm\sqrt2$ and product of moduli squared equal to $1$ .

Why this matters

This dot point ties together the whole complex numbers strand: roots, conjugates and the geometry of the Argand diagram all feed into reading and writing factorisations. It is also the bridge to partial fractions in Unit 4, where complete factorisation of a denominator is the first step.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 TASC6 marksP(z) = z^4 + alpha z^3 + beta z^2 + 18z + 45, with alpha and beta real. (a) Determine alpha and beta given that z = 3i is a root of P(z) = 0. (b) Hence solve P(z) = 0, giving all solutions in Cartesian form.

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(a) Because the coefficients are real, complex roots occur in conjugate pairs, so z = -3i is also a root. The corresponding real quadratic factor is (z - 3i)(z + 3i) = z^2 + 9. Since the constant term is 45 and 45 / 9 = 5, write P(z) = (z^2 + 9)(z^2 + az + 5). Expanding gives z^4 + a z^3 + 14 z^2 + 9a z + 45. Matching the z term: 9a = 18, so a = 2, the z^2 coefficient is 14. Hence alpha = 2 and beta = 14. (3 marks)

(b) The remaining factor z^2 + 2z + 5 = 0 gives z = (-2 +/- sqrt(4 - 20))/2 = -1 +/- 2i. The four roots are z = 3i, z = -3i, z = -1 + 2i and z = -1 - 2i. (3 marks)

2022 TASC8 marksP(z) = z^4 - 8z^3 + 33z^2 + alpha z + beta, with alpha and beta real. (a) Given one root of P(z) = 0 is z = 2 + 3i, determine alpha and beta. (b) Hence determine the other three roots of P(z) = 0.

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(a) Real coefficients force the conjugate 2 - 3i to be a root as well, giving the real quadratic factor (z - (2 + 3i))(z - (2 - 3i)) = z^2 - 4z + 13. Write P(z) = (z^2 - 4z + 13)(z^2 + az + c). Expanding: z^4 + (a - 4)z^3 + (c - 4a + 13)z^2 + (13a - 4c)z + 13c. Match z^3: a - 4 = -8, so a = -4. Match z^2: c - 4(-4) + 13 = 33, so c = 4. Then alpha = 13a - 4c = -52 - 16 = -68 and beta = 13c = 52. (4 marks)

(b) The second factor is z^2 - 4z + 4 = (z - 2)^2, giving a repeated root z = 2. The other three roots are therefore z = 2 - 3i, z = 2 and z = 2 (a double root). (4 marks)