Factorising polynomials over the complex numbers - TCE Mathematics Specialised (Tasmania)

Syllabus dot point

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

TCESpecialist MathematicsUnit 39 min readUpdated 2 Jun 2026

Quick answer

Every polynomial of degree $n$ has exactly $n$ complex roots, counting multiplicity. For a polynomial with real coefficients, non-real roots come in conjugate pairs, so it factorises into real linear factors and real quadratic factors. Over the complex numbers it factorises completely into linear factors.

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.

Fundamental theorem of algebra

Every polynomial of degree $n \ge 1$ with complex coefficients has exactly $n$ roots in the complex numbers, when each root is counted as many times as its multiplicity. Equivalently, it factorises completely into $n$ linear factors over the complex field.

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

Conjugate root theorem

If a polynomial has only real coefficients and $z = a + bi$ (with $b \ne 0$ ) is a root, then its conjugate $\bar z = a - bi$ is also a root. Non-real roots of real polynomials always occur in conjugate pairs.

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 $1$ .

Forgetting a complex root must bring its conjugate

A common error is to use a single complex root to form a linear factor and then expect the rest of the polynomial to be real. With real coefficients this never works on its own. Always pair the complex root with its conjugate to build a real quadratic factor first; only then do the remaining real factors come out cleanly.

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 questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice6 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)

Practice8 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)

Practise this

Sources & how we know this

Mathematics Specialised Level 4 (MTS415118) course document (2018 onwards) — TASC