The factor theorem is a tool for factorising polynomials: if substituting x = a into a polynomial f(x) gives zero, then (x − a) is a factor. At GCSE Higher, it is mainly used to factorise cubic expressions so that all three roots can be found.
What is the factor theorem?
The factor theorem: for a polynomial f(x), if f(a) = 0 for some value a, then (x − a) is a factor of f(x).
The converse is equally useful: if (x − a) is a factor, then f(a) = 0.
Why it works: if (x − a) divides f(x) exactly, then f(x) = (x − a) × g(x) for some polynomial g(x). Substituting x = a gives f(a) = (a − a) × g(a) = 0 × g(a) = 0. So a is a root of the equation f(x) = 0.
How do you find a factor using the factor theorem?
Step 1: Identify integer factors of the constant term in f(x). These are the only integer values of a that could make f(a) = 0.
Step 2: Test each factor by substituting it into f(x).
Step 3: The first value that gives f(a) = 0 means (x − a) is a factor.
Worked example: Show that (x − 1) is a factor of f(x) = x³ − 6x² + 11x − 6.
Test x = 1: f(1) = 1³ − 6(1²) + 11(1) − 6 = 1 − 6 + 11 − 6 = 0 ✓
Since f(1) = 0, the factor theorem confirms (x − 1) is a factor.
How do you fully factorise a cubic using the factor theorem?
Once you have found one linear factor (x − a), divide the cubic by it to obtain a quadratic. Then factorise the quadratic by the usual methods.
Full worked example: Factorise x³ − 6x² + 11x − 6 completely.
Step 1 — Find a linear factor. Constant term = −6; integer factor pairs to test: ±1, ±2, ±3, ±6.
| x | f(x) = x³ − 6x² + 11x − 6 | Result |
|---|---|---|
| 1 | 1 − 6 + 11 − 6 | 0 ✓ |
(x − 1) is a factor.
Step 2 — Divide by (x − 1).
Using polynomial long division or inspection:
x³ − 6x² + 11x − 6 = (x − 1)(x² − 5x + 6)
Verify: (x − 1)(x² − 5x + 6) = x³ − 5x² + 6x − x² + 5x − 6 = x³ − 6x² + 11x − 6 ✓
Step 3 — Factorise the quadratic.
x² − 5x + 6 = (x − 2)(x − 3) (factors of 6 that add to −5: −2 and −3)
Fully factorised: x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3)
The three roots of f(x) = 0 are x = 1, x = 2, and x = 3. You can confirm all three using the factor theorem:
f(1) = 0 ✓, f(2) = 8 − 24 + 22 − 6 = 0 ✓, f(3) = 27 − 54 + 33 − 6 = 0 ✓
How do you perform polynomial long division?
Dividing x³ − 6x² + 11x − 6 by (x − 1):
- Divide the leading term: x³ ÷ x = x². Multiply: x²(x − 1) = x³ − x². Subtract: (x³ − 6x² + 11x − 6) − (x³ − x²) = −5x² + 11x − 6.
- Divide: −5x² ÷ x = −5x. Multiply: −5x(x − 1) = −5x² + 5x. Subtract: (−5x² + 11x − 6) − (−5x² + 5x) = 6x − 6.
- Divide: 6x ÷ x = 6. Multiply: 6(x − 1) = 6x − 6. Subtract: (6x − 6) − (6x − 6) = 0.
Quotient: x² − 5x + 6, remainder 0. ✓
What should you do if the first value you test is not a root?
Keep testing the factors of the constant term systematically. If none of the small integers works, double-check your arithmetic — for a GCSE question, there will always be at least one integer root that the factor theorem reveals.
Frequently asked questions
What values of a should I try first?
Always try x = 1 and x = −1 first because they are the easiest to compute. If those fail, move to the other integer factors of the constant term (positive and negative). For x³ − 6x² + 11x − 6, the constant is −6, giving candidates ±1, ±2, ±3, ±6.
Can the factor theorem find all three roots of a cubic?
After finding the first root using the factor theorem and dividing to get a quadratic, the quadratic's two roots give the remaining two factors. You can also use the factor theorem on the quadratic, but standard factorising (or the quadratic formula) is usually faster at that stage.
Is the factor theorem on the GCSE Higher specification?
The factor theorem appears on GCSE Higher specifications for AQA, Edexcel and OCR. It is typically examined alongside algebraic long division and factorising cubic expressions. Confirm the exact syllabus year with your teacher, as the depth of coverage varies slightly between boards.
How do I check my fully factorised answer?
Expand your factorised form and verify it equals the original polynomial. For (x − 1)(x − 2)(x − 3): first expand (x − 1)(x − 2) = x² − 3x + 2, then multiply by (x − 3): x³ − 3x² − 3x² + 9x + 2x − 6 = x³ − 6x² + 11x − 6 ✓. Alternatively, check that substituting each root into the original polynomial gives zero.
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