QCE Specialist Mathematics - Unit 3 - Further complex numbers

Factorisation of polynomials

Revise Specialist Mathematics factor and remainder theorems, conjugate roots and solving polynomial equations over the complex numbers.

Part of the free QCE Specialist Mathematics notes library for Unit 3: Further complex numbers.

Updated 2026-05-18 - 5 min read

Polynomial questions in Specialist often look like algebra, but they are really about roots. A factor tells you a root, a root tells you a factor, and complex roots are allowed because the syllabus asks you to solve polynomial equations over $\mathbb C$.

Remainder and factor theorem

If a polynomial $P(x)$ is divided by $x-a$, the remainder is:

$ P(a). $

So:

$ x-a\text{ is a factor of }P(x)\quad \Longleftrightarrow \quad P(a)=0. $

That is the factor theorem. It is useful because it turns a factorisation problem into a substitution problem.

The division statement behind this is:

$ P(x)=(x-a)Q(x)+R $

where $R$ is a constant because the divisor is linear. Substituting $x=a$ removes the quotient term and leaves $P(a)=R$. That is why a remainder theorem question is usually much faster by substitution than by full polynomial division.

Complex conjugate roots

If a polynomial has real coefficients and $a+bi$ is a root, then $a-bi$ is also a root. This is the complex conjugate root theorem.

For example, if a real cubic has roots $2+i$ and $4$, then it must also have root $2-i$. The factors are:

$ (x-(2+i))(x-(2-i))(x-4). $

The conjugate pair multiplies neatly:

$ (x-2-i)(x-2+i)=(x-2)^2+1. $

The theorem comes from conjugating the equation $P(a+bi)=0$. If $P$ has real coefficients, then:

$ \overline{P(a+bi)}=P(a-bi). $

Since the conjugate of $0$ is still $0$, $P(a-bi)=0$. That is why the real-coefficient condition matters.

Solving over $\mathbb C$

Over the complex numbers, a polynomial of degree $n$ has $n$ roots counting multiplicity. In exam work, you usually combine these tools:

• use a given factor or root
• apply the conjugate root theorem if coefficients are real
• divide or factor the polynomial
• solve the remaining quadratic or linear factor

For degree $4$ questions, do not expect every root to appear immediately. A typical path is to find one factor by the factor theorem, use the conjugate root theorem when the coefficients are real, then reduce the polynomial to a quadratic. If the coefficients are imaginary, use direct factorisation, synthetic division or coefficient comparison instead of inventing a conjugate pair.

Repeated roots still count. For example, $(x-2)^2(x^2+1)=0$ has four complex roots counted with multiplicity: $2,2,i,-i$. If the question asks for factors, keep the repeated factor. If it asks for distinct solutions, list each value once.

Exam workflow

When a polynomial has a known factor, divide by that factor first. When it has a known root, convert it into a factor by writing $x-a$. When a complex root is given for a real-coefficient polynomial, immediately write down the conjugate root as well. These small translations keep the algebra controlled.

Coefficient comparison is also useful. If a monic cubic with roots $p$, $q$ and $r$ is written as:

$ (x-p)(x-q)(x-r), $

then expanding in pairs often avoids messy complex arithmetic. Conjugate pairs should be multiplied first because:

$ (x-(a+bi))(x-(a-bi))=(x-a)^2+b^2. $

This creates a real quadratic, which is exactly what you want before expanding the final polynomial.

Polynomial division details

The general division statement is:

$ P(z)=D(z)Q(z)+R(z), $

where the degree of $R(z)$ is lower than the degree of $D(z)$. If $D(z)$ is linear, the remainder is a constant. If $D(z)$ is quadratic, the remainder has the form $az+b$.

That second case is useful when the divisor factorises. Suppose $x-4$ is a factor of $P(x)$ and division by $x+1$ leaves remainder $10$. If you are asked for the remainder when dividing by:

$ x^2-3x-4=(x-4)(x+1), $

write:

$ P(x)=(x-4)(x+1)Q(x)+ax+b. $

Then $P(4)=0$ gives $4a+b=0$, and $P(-1)=10$ gives $-a+b=10$. Solving those two equations gives the linear remainder. This method is much faster than trying to build the whole polynomial.

For division by $az-b$, the remainder is:

$ P\left(\frac{b}{a}\right), $

because $az-b=0$ when $z=\frac{b}{a}$.

Long division is still worth knowing because it shows where the quotient comes from. For example, divide:

$ P(x)=x^3-2x^2-5x+6 $

by $x-3$. The first term is $x^2$ because $x^3\div x=x^2$. Multiply back and subtract:

$ x^3-2x^2-5x+6-(x^3-3x^2)=x^2-5x+6. $

The next term is $x$:

$ x^2-5x+6-(x^2-3x)=-2x+6. $

The final term is $-2$:

$ -2x+6-(-2x+6)=0. $

So:

$ x^3-2x^2-5x+6=(x-3)(x^2+x-2). $

Then:

$ x^2+x-2=(x+2)(x-1). $

The roots are $3$, $-2$ and $1$.

Real and complex factorisation

Over the complex numbers, a degree $n$ polynomial has $n$ roots counting multiplicity, so it can be written as linear factors. Over the real numbers, non-real conjugate pairs stay together as irreducible real quadratics.

For example:

$ (z-(2+i))(z-(2-i))=(z-2)^2+1. $

So a real factorisation may stop at:

$ (z-3)((z-2)^2+1), $

while the complex factorisation continues into:

$ (z-3)(z-2-i)(z-2+i). $

The wording "factorise over the real numbers" and "factorise over the complex numbers" is therefore not cosmetic; it changes the expected final form.

Guided complex factorisation

Suppose a real quartic has leading coefficient $1$ and a known root $2+i$. Then $2-i$ is also a root, so the conjugate pair contributes:

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

If the polynomial is:

$ P(x)=x^4-6x^3+14x^2-22x+15, $

divide by $x^2-4x+5$ to get:

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

The remaining quadratic has roots:

$ x=1\pm i\sqrt2. $

So the complex factorisation is:

$ (x-2-i)(x-2+i)(x-1-i\sqrt2)(x-1+i\sqrt2). $

Over the real numbers, the factorisation stops at:

$ (x^2-4x+5)(x^2-2x+3). $

This is the main workflow for complex polynomial questions: use the known complex root, add the conjugate if coefficients are real, divide by the real quadratic, then solve what remains.

Quick check

Sources

• QCAA Specialist Mathematics subject page
• QCAA Specialist Mathematics 2025 syllabus
• OpenStax College Algebra 2e: Zeros of Polynomial Functions
• OpenStax Algebra and Trigonometry 2e: Polar Form of Complex Numbers

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