Complex numbers as roots of polynomials

IB Mathematics: Applications and Interpretation HL· 1.13· 15 min read

1. The Conjugate Root Theorem★★☆☆☆⏱ 5 min

For any polynomial with only real coefficients, if a non-real complex number $z = a + bi$ is a root, then its complex conjugate $\overline{z} = a - bi$ must also be a root. This means non-real complex roots always occur in matched conjugate pairs for polynomials with real coefficients.

📘 Definition

Conjugate Root Theorem

For $P (x)$ with real coefficients

If $a + bi$ is a root of $P (x) = 0$ , then $a - bi$ is also a root of the equation

Example:

If $2 + 3 i$ is a root, $2 - 3 i$ is guaranteed to also be a root

📐 Worked Example

The cubic polynomial $P (x) = x^{3} - 5 x^{2} + 7 x + 13$ has a root at $x = 2 - 3 i$ . Find the other two roots.

1
By the conjugate root theorem, since $P (x)$ has real coefficients, the conjugate of $2 - 3 i$ is also a root.
2
Multiply the linear factors for the two complex roots:
3
$(x - (2 - 3i))(x - (2 + 3i)) = ((x-2) + 3i)((x-2) - 3i) = (x-2)^2 - (3i)^2 = x^2 - 4x + 13$
4
Factor $P (x)$ as $(x - k) (x^{2} - 4 x + 13)$ , where $k$ is the third real root. Expand and equate coefficients:
5
$(x -k)(x^2 -4x +13) = x^3 -(4 +k)x^2 + (13 +4k)x -13k$
6
Equate constant terms: $- 13 k = 13 ⟹ k = - 1$ . Check the $x^{2}$ coefficient: $- (4 + (- 1)) = - 5$ , which matches the original polynomial.
7
The other two roots are therefore:

2. Finding All Roots of a Polynomial★★★☆☆⏱ 5 min

When given one non-real complex root of a polynomial with real coefficients, you can use the conjugate root theorem to get a second root, form a quadratic factor from the two roots, divide the original polynomial by this quadratic, and solve the remaining quotient to get any remaining roots.

✓ Quick check

Test your understanding of the core rule:

A quartic (degree 4) polynomial with real coefficients has one known non-real complex root. How many non-real complex roots does it have in total?
- 2
- 3
- 4
- Cannot tell
Reveal answer
2 —
Correct! Non-real complex roots come in pairs, so there are 2 non-real roots total, and the other two roots are real.

📐 Worked Example

Find all roots of $Q (x) = x^{4} - 6 x^{3} + 20 x^{2} - 24 x + 16$ , given $x = 1 + 3 i$ is a root.

1
By the conjugate root theorem, $1 - 3 i$ is also a root.
2
Multiply the linear factors for the two complex roots to get a quadratic factor:
3
$(x - (1+\sqrt{3}i))(x - (1 - \sqrt{3}i)) = (x-1)^2 - (\sqrt{3}i)^2 = x^2 - 2x + 4$
4
Divide $Q (x)$ by $x^{2} - 2 x + 4$ to get the remaining quadratic factor:
5
$x^4 - 6x^3 + 20x^2 - 24x + 16 = (x^2 - 2x + 4)(x^2 - 4x + 4)$
6
Factor the remaining quadratic: $x^{2} - 4 x + 4 = (x - 2)^{2}$
7
All roots are: $1 + 3 i$ , $1 - 3 i$ , and $2$ (with multiplicity 2).

3. Constructing Polynomials From Given Roots★★★☆☆⏱ 5 min

If you are given a set of roots including one or more non-real complex numbers for a polynomial with real coefficients, you first add the conjugate of each non-real root to your list of roots, then multiply all corresponding linear factors to get the full polynomial.

📐 Worked Example

Find a monic (leading coefficient 1) cubic polynomial with real coefficients that has roots at $3$ and $1 - 2 i$ . Write your answer in expanded form.

1
Since the polynomial has real coefficients and a non-real root $1 - 2 i$ , $1 + 2 i$ must also be a root. We now have all three roots: $3$ , $1 - 2 i$ , $1 + 2 i$ .
2
Write the polynomial in factored form:
3
$P(x) = (x - 3)(x - (1 - 2i))(x - (1 + 2i))$
4
Multiply the factors for the complex roots first:
5
$(x - 1 + 2i)(x - 1 - 2i) = (x-1)^2 - (2i)^2 = x^2 - 2x + 1 + 4 = x^2 - 2x + 5$
6
Multiply by the linear factor from the real root:
7
$(x - 3)(x^2 - 2x + 5) = x^3 - 2x^2 + 5x - 3x^2 + 6x - 15 = x^3 - 5x^2 + 11x - 15$
8
This is the required monic cubic polynomial.

4. Common Pitfalls

Wrong move:

Applying the conjugate root theorem to polynomials with complex coefficients.

Why:

The theorem only guarantees conjugate pairs for polynomials with all real coefficients.

Correct move:

Only use the conjugate root theorem when the polynomial is stated to have real coefficients, which is always the case in IB AI HL problems.

Wrong move:

Making a sign error when expanding $(x - (a + bi)) (x - (a - bi))$ , ending with $- b^{2}$ instead of $+ b^{2}$ .

Why:

$(bi)^{2} = - b^{2}$ , so $- (bi)^{2} = + b^{2}$ , a common sign mistake that changes the entire polynomial.

Correct move:

Memorize the simplified form: $(x - (a + bi)) (x - (a - bi)) = (x - a)^{2} + b^{2}$ , always with a plus sign on $b^{2}$ .

Wrong move:

Claiming all roots of an odd degree polynomial with real coefficients must be real.

Why:

An odd degree polynomial only needs at least one real root; the remaining even number of roots can be non-real conjugate pairs.

Correct move:

Expect at least one real root for an odd degree polynomial, but do not assume all roots are real.

Wrong move:

Stopping after finding the conjugate of the given complex root, forgetting to check for repeated roots.

Why:

Roots can repeat, even for complex roots, so you must fully factor the entire polynomial to get all roots.

Correct move:

After dividing by the quadratic from the conjugate pair, always fully factor the remaining quotient to account for repeated roots.

5. Quick Reference Cheatsheet

Rule	Application/Formula
Conjugate Root Theorem (real coefficients)	If $a + bi$ is a root, $a - bi$ is also a root
Product of conjugate factors	$(x - (a + bi)) (x - (a - bi)) = (x - a)^{2} + b^{2}$
Odd degree polynomial (real coefficients)	At least 1 real root, rest non-real in pairs
Even degree polynomial (real coefficients)	Can have 0 real roots, all non-real in pairs
Total number of roots	Degree $n$ = $n$ roots counting multiplicity

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · 1
Find all roots given one complex root
2023 · 1
Construct polynomial from complex roots

Going deeper

syllabusIB AI HL Official SyllabusCheck section 1.13 for full requirements

What's Next

This topic is a foundational algebraic skill that you will use throughout the IB AI HL course. You will apply the conjugate root theorem to find roots of characteristic equations for linear recurrence relations and second-order differential equations later in the course, and it is a regularly tested topic in Paper 1. Mastery of this concept reinforces your understanding of polynomial structure and the properties of real and complex numbers, which supports all other topics involving polynomials and equation solving.