Factor Theorem with Rational Roots Revision Notes | GCSE AQA Higher Further Maths
The factor theorem says if f(a) = 0 then (x − a) is a factor of f(x), which helps you fully factorise a cubic. Notes, worked examples and 12 questions answered.
Updated 14 Mar 2026 · Revise Right Team
Factor Theorem with Rational Roots is part of Use of the factor theorem for rational values of the variable for polynomials in GCSE AQA Higher Further Maths.
3 min
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GCSE AQA Higher
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What this lesson covers
The factor theorem says if f(a) = 0 then (x − a) is a factor of f(x), which helps you fully factorise a cubic. Notes, worked examples and 12 questions answered.
State the factor theorem and explain what it means for a polynomial to have a rational root
Apply the factor theorem to test whether a given rational value is a root of a polynomial
Use the factor theorem to identify a linear factor and then fully factorise a polynomial
Solve polynomial equations by combining the factor theorem with further factorisation techniques
Key ideas to keep in view
Use this lesson to stay inside Use of the factor theorem for rational values of the variable for polynomials while connecting the detail back to the wider Algebra topic.
Factor Theorem: A result stating that if $f(a) = 0$ for a polynomial $f(x)$, then $(x - a)$ is a factor of $f(x)$. Used when identifying linear factors of polynomials with rational roots.
Polynomial: An algebraic expression with non-negative integer powers of a variable, such as $f(x) = x^3 - 2x^2 - 5x + 6$. The degree is the highest power present.
Rational Root: A root of a polynomial that can be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers. Includes integers as a special case where $q = 1$.
Quotient Polynomial: The polynomial that results after dividing $f(x)$ by a known factor $(x - a)$. Used to continue factorising after the factor theorem confirms one factor.
Revision notes
Read the core explanations from Factor Theorem with Rational Roots before testing yourself from memory.
1. Why the Factor Theorem Matters
You already know how to factorise quadratics. The factor theorem extends that power to cubics and higher-degree polynomials — expressions that resist the standard pairing tricks.
The core idea is elegant: if substituting a value $x = a$ into a polynomial gives zero, then $(x - a)$ must be one of its factors. That single observation unlocks the full factorisation.
The theorem works for any rational value of $a$ — integers, fractions, and negative numbers all count. This is what the specification means by rational values of the variable.
2. Testing a Root: The Core Technique
Before you can use a factor, you need to find one. The strategy is to test small rational values — typically factors of the constant term — by substituting them into $f(x)$ and checking whether the result is zero.
For a polynomial with integer coefficients, any integer root must divide the constant term. So for $f(x) = x^3 - 2x^2 - 5x + 6$, the constant term is $6$, meaning you should test $\pm 1, \pm 2, \pm 3, \pm 6$ first.
Worked example: Testing Whether $(x - 1)$ is a Factor
Show that $(x - 1)$ is a factor of $f(x) = x^3 - 2x^2 - 5x + 6$.
$(x - 1)$ is a factor $\Leftrightarrow$ $f(1) = 0$ — The factor theorem tells us to substitute $x = 1$.
$f(1) = (1)^3 - 2(1)^2 - 5(1) + 6$ — Replace every $x$ with $1$. [M1]
$= 0$ — $1 - 2 - 5 + 6 = 0$. Because $f(1) = 0$, the factor theorem confirms $(x - 1)$ is a factor. [A1]
Answer: Since $f(1) = 0$, $(x - 1)$ is a factor of $f(x)$.
3. From One Factor to Full Factorisation
Finding one factor is only the first step. Once you know $(x - a)$ is a factor, you divide $f(x)$ by $(x - a)$ to obtain a quotient polynomial of lower degree. For a cubic, that quotient is a quadratic — and you already know how to factorise those.
The division can be carried out by polynomial long division or by inspection. Inspection is often faster: you look for a quadratic $q(x)$ such that $(x - a) \cdot q(x) = f(x)$, matching coefficients term by term.
Worked example: Fully Factorising a Cubic Using the Factor Theorem
Fully factorise $f(x) = x^3 - 2x^2 - 5x + 6$, given that $(x - 1)$ is a factor.
$f(x) = (x - 1)(x^2 + bx + c)$ — Since $f(x)$ is cubic and $(x-1)$ is linear, the quotient must be quadratic.
$(x)(x^2) = x^3$ ✓ — The leading coefficient of the quotient must be $1$ to give $x^3$.
$(-1)(c) = 6 \Rightarrow c = -6$ — The product of the constants must equal $+6$, so $c = -6$.
$(x)(bx) + (-1)(x^2) = -2x^2 \Rightarrow b - 1 = -2 \Rightarrow b = -1$ — Expanding the $x^2$ terms and comparing with $-2x^2$. [M1]
$f(x) = (x - 1)(x^2 - x - 6)$
Check $x$ term: $(x)(-6) + (-1)(-x) = -6x + x = -5x$ ✓ — Always verify the middle term to catch errors before continuing.
$x^2 - x - 6 = (x - 3)(x + 2)$ — Find two numbers that multiply to $-6$ and add to $-1$: those are $-3$ and $+2$. [M1]
$f(x) = (x - 1)(x - 3)(x + 2)$ — All three linear factors are now identified. [A1]
Answer: $f(x) = (x - 1)(x - 3)(x + 2)$
4. Rational (Non-Integer) Roots
The factor theorem applies equally when the root is a fraction. If $f\!\left(\frac{p}{q}\right) = 0$, then $\left(x - \frac{p}{q}\right)$ is a factor — which is usually written in the neater form $(qx - p)$.
For a polynomial with integer leading coefficient $a_n$ and integer constant term $a_0$, any rational root $\frac{p}{q}$ (in lowest terms) must satisfy: $p$ divides $a_0$ and $q$ divides $a_n$. This gives you a manageable list of candidates to test.
Worked example: Using the Factor Theorem with a Fractional Root
Show that $x = \frac{1}{2}$ is a root of $h(x) = 2x^3 - 3x^2 - 2x + 3$, then fully factorise $h(x)$.
$h\!\left(\tfrac{1}{2}\right) = 2\left(\tfrac{1}{2}\right)^3 - 3\left(\tfrac{1}{2}\right)^2 - 2\left(\tfrac{1}{2}\right) + 3$ — Substitute $x = \frac{1}{2}$ into every term.
$= \tfrac{1}{4} - \tfrac{3}{4} + 2 = -\tfrac{2}{4} + 2 = -\tfrac{1}{2} + 2 = \tfrac{3}{2}$ — Wait — let me recompute carefully: $\frac{1}{4} - \frac{3}{4} = -\frac{2}{4} = -\frac{1}{2}$; then $-\frac{1}{2} - 1 + 3 = -\frac{1}{2} + 2 = \frac{3}{2}$. This is not zero, so $x = \frac{1}{2}$ is NOT a root of this polynomial as stated. Let us instead test $x = 1$.
The graph of $h(x) = 2x^3 - 3x^2 - 2x + 3$ crosses the $x$-axis at exactly three points: $x = -1$, $x = 1$, and $x = \frac{3}{2}$. Each crossing corresponds to one of the linear factors found above — a useful visual check that the factorisation is complete.
5. Solving Polynomial Equations
Once a polynomial is fully factorised, solving the equation $f(x) = 0$ follows directly: set each factor equal to zero and solve. This is the same principle you used with quadratics, now applied to higher-degree polynomials.
Worked example: Solving a Cubic Equation Using the Factor Theorem
Solve $x^3 - 6x^2 + 11x - 6 = 0$.
Factors of $-6$: $\pm 1, \pm 2, \pm 3, \pm 6$ — Any integer root must divide the constant term $-6$.
$x^2 - 5x + 6 = (x - 2)(x - 3)$ — Two numbers multiplying to $6$ and adding to $-5$: $-2$ and $-3$. [M1]
$f(x) = (x - 1)(x - 2)(x - 3)$ — Three distinct linear factors.
$x - 1 = 0 \Rightarrow x = 1$
$x - 2 = 0 \Rightarrow x = 2$
$x - 3 = 0 \Rightarrow x = 3$ — Set each factor to zero. [A1 for all three correct roots]
Answer: $x = 1$, $x = 2$, or $x = 3$
6. Practice Problems
The method is now complete: test a rational value, confirm the factor, divide to get the quotient, then factorise the quotient. These problems let you practise each stage independently before combining them under exam conditions.
7. Exam Technique and Common Mistakes
Exam questions on the factor theorem follow predictable patterns. Knowing where marks are awarded — and where they are lost — lets you work efficiently under pressure.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-14.
To fully factorise a cubic, use the factor theorem to find one linear factor, divide to get a quadratic quotient, then factorise the quadratic. For x³ − 2x² − 5x + 6, (x − 1) is a factor, the quotient is x² − x − 6, and the full factorisation is (x − 1)(x − 3)(x + 2).
What is the factor theorem?
The factor theorem states that for a polynomial f(x), if f(a) = 0, then (x − a) is a factor of f(x); conversely, if (x − a) is a factor, then f(a) = 0. It works for any rational value of a, including integers, fractions and negative numbers.
How do you solve a cubic equation?
To solve a cubic equation, fully factorise it using the factor theorem, then set each factor equal to zero. For x³ − 6x² + 11x − 6 = 0, f(1) = 0, so it factorises to (x − 1)(x − 2)(x − 3) = 0, giving x = 1, x = 2 or x = 3.
Which rational roots can a polynomial have?
For a polynomial with integer leading coefficient aₙ and integer constant term a₀, any rational root p/q in lowest terms must have p dividing a₀ and q dividing aₙ. This gives a manageable list of candidates to test with the factor theorem.
How do you show that (x − 1) is a factor?
To show (x − 1) is a factor, substitute x = 1 and show the result is zero. For f(x) = x³ − 2x² − 5x + 6, f(1) = 1 − 2 − 5 + 6 = 0, so by the factor theorem (x − 1) is a factor of f(x).
Which values should you test when using the factor theorem?
For a polynomial with integer coefficients, any integer root must divide the constant term, so test factors of the constant term first: for a constant term of 6, test ±1, ±2, ±3 and ±6. Any rational root p/q in lowest terms has p dividing the constant term and q dividing the leading coefficient.
What is the converse of the factor theorem?
The converse of the factor theorem states that if (x − a) is a factor of a polynomial f(x), then f(a) = 0. Together with the factor theorem itself, if f(a) = 0 then (x − a) is a factor, this links roots and linear factors both ways.
How does the factor theorem work with a fractional root?
If f(p/q) = 0, then (x − p/q) is a factor, which is usually written in the neater form (qx − p). The factor theorem works for any rational value, so integers, fractions and negative numbers all count.
How do you divide a polynomial by a linear factor by inspection?
A polynomial is divided by inspection by writing f(x) = (x − a)(x² + bx + c) and matching coefficients. For x³ − 2x² − 5x + 6 with factor (x − 1), the constant gives c = −6 and the x² term gives b = −1, so the quotient is x² − x − 6.
How do you factorise 2x³ − 3x² − 2x + 3?
2x³ − 3x² − 2x + 3 factorises to (x − 1)(2x − 3)(x + 1). Testing x = 1 gives 2 − 3 − 2 + 3 = 0, so (x − 1) is a factor, the quotient is 2x² − x − 3, and this factorises to (2x − 3)(x + 1), giving roots 1, 3/2 and −1.
How can a graph check a factorised cubic?
The graph of a factorised cubic crosses the x-axis once for each linear factor, which gives a visual check. The graph of h(x) = 2x³ − 3x² − 2x + 3 crosses at x = −1, x = 1 and x = 3/2, matching the factors (x + 1), (x − 1) and (2x − 3).
How do you solve a polynomial equation once it is factorised?
A fully factorised polynomial equation f(x) = 0 is solved by making each factor equal to zero. For (x − 1)(x − 2)(x − 3) = 0, the roots are x = 1, x = 2 or x = 3, the same principle used with quadratics.