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

§-Syllabus dot point

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 4.89 min answerUpdated 2026-06-02

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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

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.

Step 1: Apply the conjugate root theorem to find the missing root

Because the polynomial has real coefficients, any non-real complex root must be accompanied by its conjugate. This guarantees we have identified all three roots of the cubic before we start building factors.

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

Step 2: Form the real quadratic factor from the conjugate pair

Multiplying the two conjugate linear factors together always produces a quadratic with real coefficients, which is what we want so that the final polynomial has no complex arithmetic left in its coefficients.

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

Step 3: Multiply by the remaining linear factor and expand

Now that we have a real quadratic and one more real root, we form the full cubic by multiplying these together and collecting like terms.

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

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 $x^2$ .

Final answer: $P(x) = x^3 - 9x^2 + 25x - 25$ .

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.

TCE 20236 marks

P(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.

Show worked answer →

(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 $\dfrac{45}{9} = 5$ , 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 $a = 2$ , and the $z^2$ coefficient is $14$ . Hence $\alpha = 2$ and $\beta = 14$ . (3 marks)

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

TCE 20228 marks

P(z) = z^4 - 8z^3 + 33z^2 + \alpha z + \beta

, with

\alpha

and

\beta

real. (a) Given one root of

P(z) = 0

z = 2 + 3i

, determine

\alpha

and

\beta

. (b) Hence determine the other three roots of

P(z) = 0

Show worked answer →

(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 gives $z^4 + (a - 4)z^3 + (c - 4a + 13)z^2 + (13a - 4c)z + 13c$ . 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)