QCE Specialist Mathematics - Unit 3 - Further complex numbers

Factorisation of Polynomials | QCE Specialist Mathematics

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