The remainder theorem says that when f(x) is divided by (x − a), the remainder is f(a). The factor theorem is the special case where that remainder is zero: (x − a) is a factor of f(x) exactly when f(a) = 0. Together they let you find remainders, test possible factors, find unknown coefficients and factorise cubics without long division.
The factor and remainder theorems
- Remainder theorem: dividing f(x) by (x − a) leaves remainder f(a)
- Dividing by (ax − b) leaves remainder f(b/a)
- Factor theorem: (x − a) is a factor of f(x) if and only if f(a) = 0
- To find a factor of a cubic, try x = ±(factors of the constant term)
The remainder theorem
To find the remainder when f(x) is divided by a linear expression, find the value of x that makes the divisor zero and substitute it into f(x). For (x − a), substitute x = a; for (2x − 1), substitute x = ½. There is no need to carry out the division.
The factor theorem
If the remainder is zero, the divisor divides exactly, so it is a factor. So to test whether (x − a) is a factor, work out f(a): if it equals 0, (x − a) is a factor. For (ax − b), test f(b/a).
In a “show that” question, state the conclusion clearly: “f(3) = 0, so (x − 3) is a factor”.
Finding unknown coefficients
Each factor or remainder condition gives one equation. A stated factor (x − a) gives f(a) = 0; a stated remainder r gives f(a) = r. With two unknown coefficients you need two conditions, then solve the equations simultaneously.
Factorising a cubic fully
Find one factor by trying small values of x, especially the positive and negative factors of the constant term, until f(a) = 0. Then write f(x) = (x − a)(x² + bx + c) and find b and c by comparing coefficients, or by long division. Finally, factorise the quadratic.
Comparing coefficients is usually quickest: match the x² terms and the constant terms.
Solving cubics
To solve f(x) = 0, factorise fully and set each factor equal to zero. If the remaining quadratic does not factorise, use the formula and give exact surd roots. Watch for a fractional root from a factor such as (2x − 1).
Worked examples
Example 1. Find the remainder when f(x) = x³ − 2x² + 5x − 1 is divided by (x − 2).
- x − 2 = 0 at x = 2.
- f(2) = 8 − 8 + 10 − 1.
Answer: The remainder is 9.
Marks: M1 for f(2), A1 for 9.
Example 2. Show that (x − 2) is a factor of f(x) = x³ + 2x² − 5x − 6.
- Test f(2), since x − 2 = 0 at x = 2.
- f(2) = 8 + 8 − 10 − 6 = 0.
Answer: f(2) = 0, so (x − 2) is a factor.
Marks: M1 for f(2), A1 for 0 and the conclusion.
Example 3. f(x) = 2x³ + ax² + bx − 4 has (x + 1) as a factor, and leaves remainder 24 when divided by (x − 2). Find a and b.
- Factor: f(−1) = 0 gives −2 + a − b − 4 = 0, so a − b = 6.
- Remainder: f(2) = 24 gives 16 + 4a + 2b − 4 = 24, so 2a + b = 6.
- Add the equations: 3a = 12, so a = 4 and b = −2.
Answer: a = 4, b = −2.
Marks: M1 for each equation, A1 for both values.
Example 4. Factorise f(x) = x³ + 2x² − 5x − 6 completely.
- f(2) = 0, so (x − 2) is a factor.
- Write f(x) = (x − 2)(x² + bx + c). x² terms: b − 2 = 2, so b = 4. Constants: −2c = −6, so c = 3.
- x² + 4x + 3 = (x + 1)(x + 3).
Answer: f(x) = (x − 2)(x + 1)(x + 3).
Marks: M1 for the factor, M1 for comparing coefficients, A1 for the full factorisation.
Example 5. Solve x³ + 2x² − 5x − 6 = 0.
- From the factorisation: (x − 2)(x + 1)(x + 3) = 0.
- Set each factor to zero.
Answer: x = 2, x = −1 or x = −3.
Marks: M1 for setting each factor to zero, A1 for all three roots.
Common mistakes
- Substituting the wrong sign. For (x + 1), substitute x = −1, not x = 1.
- Using f(b) for (ax − b). For (2x − 1), substitute x = ½.
- Confusing factor and remainder conditions. A factor gives f(a) = 0; a remainder r gives f(a) = r.
- Stopping after one factor. Divide or compare coefficients to find the quadratic, then factorise it too.
- Arithmetic slips with negative values. Use brackets when substituting: (−1)³ = −1.
How the factor and remainder theorems are examined
The factor and remainder theorems are tested at the start of Pure 2 (WMA12) for Edexcel International A-Level, and in Year 1 Pure for Edexcel A-Level (9MA0). Questions usually:
- ask you to show that a given linear expression is a factor
- find unknown coefficients from factor and remainder conditions
- factorise a cubic completely, often “hence”
- solve a cubic, giving any irrational roots exactly
Practise with the Pure 2 algebraic methods exam questions, where these questions come with mark schemes, and with the Pure 2 algebraic methods worksheet. The A-Level algebra revision engine gives unlimited further questions.
The theorems are a shortcut alongside algebraic long division, which finds the full quotient.
Frequently asked questions
What is the factor theorem?
(x − a) is a factor of f(x) if and only if f(a) = 0.
What is the remainder theorem?
When f(x) is divided by (x − a), the remainder is f(a).
How do I find a factor of a cubic?
Try x = 1, −1, 2, −2 and so on, starting with the factors of the constant term, until f(x) = 0. The corresponding (x − a) is a factor.