Solving polynomial equations over C: VCE Specialist Mathematics Unit 3

Syllabus dot point

“Solution of polynomial equations over the complex numbers, the fundamental theorem of algebra, the conjugate root theorem for polynomials with real coefficients, and the full factorisation of real polynomials into linear and irreducible quadratic factors”

6 min readUpdated 29 May 2026

Quick answer

The fundamental theorem of algebra says a degree- $n$ polynomial has exactly $n$ roots in $\mathbb{C}$ , counted with multiplicity, so it factorises into $n$ linear factors over $\mathbb{C}$ . For a polynomial with real coefficients, the conjugate root theorem says non-real roots occur in conjugate pairs: if $a + bi$ is a root then so is $a - bi$ . Multiplying a conjugate pair gives a real irreducible quadratic $z^2 - 2az + (a^2 + b^2)$ , so every real polynomial factorises over $\mathbb{R}$ into real linear factors and irreducible real quadratics. Use the factor theorem and division to find the remaining roots once one is known.

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What this dot point is asking
How many roots, and where
The conjugate root theorem
Factorising over the reals
Examples in context
Try this
Exam-style questions

What this dot point is asking

VCAA wants you to solve polynomial equations completely over $\mathbb{C}$ , using the fundamental theorem of algebra to know how many roots to expect, the conjugate root theorem to pair complex roots of real polynomials, and the factor theorem to factorise. The goal is a full factorisation into linear factors over $\mathbb{C}$ , or into real linear and irreducible quadratic factors over $\mathbb{R}$ .

How many roots, and where

The fundamental theorem of algebra guarantees that any polynomial $P(z)$ of degree $n \ge 1$ with complex coefficients has at least one complex root, and by repeated factoring exactly $n$ roots counted with multiplicity. Over $\mathbb{C}$ it therefore splits completely:

P(z) = a_n (z - z_1)(z - z_2)\cdots(z - z_n).

Some of the $z_k$ may coincide (repeated roots) or be real; the count of $n$ always holds.

The conjugate root theorem

If $P$ has real coefficients and $a + bi$ (with $b \ne 0$ ) is a root, then its conjugate $a - bi$ is also a root. The reason is that conjugation distributes over sums and products, and real coefficients are their own conjugates, so $\overline{P(z)} = P(\overline{z})$ ; if $P(z) = 0$ then $P(\overline{z}) = 0$ too.

A consequence is parity: complex roots come in pairs, so a real polynomial of odd degree must have at least one real root. A real cubic, for example, has either three real roots or one real root and one conjugate pair.

Factorising over the reals

Multiplying a conjugate pair clears the imaginary part:

(z - (a + bi))(z - (a - bi)) = z^2 - 2az + (a^2 + b^2),

a quadratic with real coefficients and negative discriminant, hence irreducible over $\mathbb{R}$ . Therefore every real polynomial factorises over $\mathbb{R}$ into real linear factors (one per real root) and irreducible real quadratics (one per conjugate pair). To carry this out:

Find one root, by inspection, the rational root test, or a given clue.
Divide $P(z)$ by the corresponding factor to reduce the degree.
Solve the reduced polynomial, using the conjugate pair to write a quadratic factor directly when a complex root is found.

Worked example

Solving a real cubic

Solve $z^3 - 5z^2 + 11z - 15 = 0$ over $\mathbb{C}$ , given that $z = 3$ is a root.

Step 1: Use the factor theorem to reduce the degree

Because $z = 3$ is a root, the factor theorem guarantees that $(z - 3)$ divides the polynomial exactly. Dividing gives:

z^3 - 5z^2 + 11z - 15 = (z - 3)(z^2 - 2z + 5).

Check by expanding: $(z - 3)(z^2 - 2z + 5) = z^3 - 2z^2 + 5z - 3z^2 + 6z - 15 = z^3 - 5z^2 + 11z - 15$ . Correct.

Step 2: Solve the remaining quadratic factor

Apply the quadratic formula to $z^2 - 2z + 5 = 0$ . The discriminant is $4 - 20 = -16 < 0$ , so the roots are complex:

z = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i.

Step 3: State all roots and verify the conjugate-pair structure

The three roots are $z = 3$ , $z = 1 + 2i$ and $z = 1 - 2i$ . Because the polynomial has real coefficients, the conjugate root theorem requires any non-real root to appear with its conjugate, and indeed $1 + 2i$ and $1 - 2i$ form a conjugate pair. Their quadratic factor $z^2 - 2z + 5$ matches the real form $z^2 - 2az + (a^2 + b^2)$ with $a = 1$ , $b = 2$ (since $a^2 + b^2 = 1 + 4 = 5$ ), confirming the factorisation is complete.

Common exam traps

Pairing conjugates for complex coefficients: The conjugate root theorem requires real coefficients. If the polynomial has complex coefficients, roots need not pair up.
Stopping short of $n$ roots: A degree- $n$ polynomial has exactly $n$ roots over $\mathbb{C}$ . Find all of them, counting multiplicity.
Wrong quadratic factor: From roots $a \pm bi$ the real factor is $z^2 - 2az + (a^2 + b^2)$ , not $z^2 + 2az + \dots$ . Watch the sign.
Assuming all roots are complex: A real odd-degree polynomial must have a real root. Try small integer values first.

Examples in context

Example 1. A real polynomial with roots $2i$ and $-2i$ has the factor $z^2 + 4$ (here $a = 0$ , $b = 2$ ).

Example 2. $z^4 + 1 = 0$ has four roots, the fourth roots of $-1$ , forming two conjugate pairs and factorising over $\mathbb{R}$ into two irreducible quadratics.

Try this

Q1. A real quartic has roots $1 + i$ and $-3$ . Name one further root it must have. [1 mark]

Cue. $1 - i$ , the conjugate of $1 + i$ .

Q2. Solve $z^2 + 4z + 13 = 0$ over $\mathbb{C}$ . [2 marks]

Cue. $z = -2 \pm 3i$ .

Q3. Given $z = 1 - i$ is a root of $z^3 - 3z^2 + 4z - 2 = 0$ , find all roots. [3 marks]

Cue. Conjugate $1 + i$ gives factor $z^2 - 2z + 2$ ; division leaves $z - 1$ , so roots are $1$ , $1 \pm i$ .

Exam-style questions

Questions in the style of VCAA 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.

2025 VCAA-style4 marks

Consider the function with rule f(z) = z^4 + 6z^2 + 25, where z is in C. Given that 1 + 2i is a solution of f(z) = 0, find a quadratic factor of f(z) and hence find all remaining solutions of f(z) = 0.

Show worked answer →

The coefficients of f are real, so by the conjugate root theorem the conjugate 1 - 2i is also a solution.

A quadratic factor is therefore (z - (1 + 2i))(z - (1 - 2i)). Expand using (z - a)(z - a-conjugate) = z^2 - 2Re(a)z + |a|^2: here 2Re = 2 and |1 + 2i|^2 = 1 + 4 = 5, so the factor is z^2 - 2z + 5.

Divide f(z) by this factor: z^4 + 6z^2 + 25 = (z^2 - 2z + 5)(z^2 + 2z + 5). (Check: the cross terms in z^3 and z cancel, and the z^2 terms give 5 - 4 + 5 = 6.)

Set the second factor to zero: z^2 + 2z + 5 = 0. Using the quadratic formula, z = (-2 +/- sqrt(4 - 20)) / 2 = (-2 +/- sqrt(-16)) / 2 = -1 +/- 2i.

So the remaining solutions are z = -1 + 2i and z = -1 - 2i (the full solution set is 1 + 2i, 1 - 2i, -1 + 2i, -1 - 2i).

2025 VCAA-style1 mark

The equation z^3 + az^2 + bz - 52 = 0, where a, b are in R and z is in C, has a solution z = 2 - 3i. The value of ab is A. -232 B. -64 C. -8 D. 0

Show worked answer →

The answer is A.

Because a and b are real, the conjugate root theorem gives a second root z = 2 + 3i. The quadratic factor from this pair is (z - (2 - 3i))(z - (2 + 3i)) = z^2 - 4z + 13.

The cubic has a third (real) root r. For z^3 + az^2 + bz - 52, the product of the three roots equals -(-52) = 52. The product of the complex pair is 13, so 13r = 52, giving r = 4.

Expand (z^2 - 4z + 13)(z - 4) = z^3 - 8z^2 + 29z - 52. Comparing coefficients, a = -8 and b = 29.

Therefore ab = (-8)(29) = -232, which is option A.