To factorise ax² + bx + c when a ≠ 1, multiply a × c, then find two integers that multiply to ac and add to b. Rewrite the middle term using those two numbers and factorise by grouping. The ac method works for every quadratic with a leading coefficient greater than 1.
What makes harder quadratics different from simpler ones?
When a = 1, factorising x² + bx + c only requires two numbers that multiply to c and add to b, giving (x + p)(x + q). When a ≠ 1 — for example 3x² + 10x + 3 — simple inspection is harder because the leading coefficient distributes across both bracket pairs. The ac method removes the guesswork by reducing the problem to a structured factor search.
What is the ac method and how does it work?
The ac method has four steps:
- Find ac: multiply the coefficient of x² (a) by the constant term (c).
- Find the factor pair: find two integers whose product is ac and whose sum is b.
- Split the middle term: rewrite bx as the sum of two x-terms using the two integers found.
- Factorise by grouping: group the four terms into two pairs and take out a common factor from each pair.
Worked example: 2x² + 7x + 3
| Step | Working |
|---|---|
| 1. Find ac | a = 2, c = 3 → ac = 6 |
| 2. Factor pair for 6 that sums to 7 | 1 × 6 = 6, 1 + 6 = 7 ✓ |
| 3. Rewrite the middle term | 2x² + 1x + 6x + 3 |
| 4a. Group | (2x² + x) + (6x + 3) |
| 4b. Factor each group | x(2x + 1) + 3(2x + 1) |
| 4c. Final factorisation | (2x + 1)(x + 3) |
Verify by expanding: (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓
Worked example: 6x² − x − 2
| Step | Working |
|---|---|
| 1. Find ac | a = 6, c = −2 → ac = −12 |
| 2. Factor pair for −12 that sums to −1 | 3 × (−4) = −12, 3 + (−4) = −1 ✓ |
| 3. Rewrite the middle term | 6x² + 3x − 4x − 2 |
| 4a. Group | (6x² + 3x) + (−4x − 2) |
| 4b. Factor each group | 3x(2x + 1) − 2(2x + 1) |
| 4c. Final factorisation | (2x + 1)(3x − 2) |
Verify: (2x + 1)(3x − 2) = 6x² − 4x + 3x − 2 = 6x² − x − 2 ✓
How do you find the right factor pair when ac is large?
Work systematically: list factor pairs of ac, keeping track of both positive and negative options since c can be negative.
Example: find a factor pair for ac = −30 that sums to −1.
| Pair | Product | Sum |
|---|---|---|
| 1, −30 | −30 | −29 |
| 2, −15 | −30 | −13 |
| 3, −10 | −30 | −7 |
| 5, −6 | −30 | −1 ✓ |
The numbers 5 and −6 give the required sum. Positive and negative pairs are equally valid candidates when c < 0.
How do you check your factorisation is correct?
Always expand the result. Multiply the two brackets using FOIL (First, Outside, Inside, Last) and confirm you recover the original quadratic. If any term does not match, go back and recheck the factor pair or the grouping step.
Additionally, substitute a simple value of x (for example x = 0 or x = 1) into both the original and the factorised form. If the values agree, the factorisation is almost certainly correct.
Frequently asked questions
What if the quadratic cannot be factorised with integers?
If no integer factor pair of ac sums to b, the quadratic does not factorise over the integers. You then solve using the quadratic formula or completing the square instead. A quick way to check is to calculate the discriminant b² − 4ac: if it is not a perfect square, the quadratic has irrational roots and cannot be factorised with integers.
Does the order in which I split the middle term matter?
No. Writing bx as either p·x + q·x or q·x + p·x both lead to the same factorised form, although the grouping step may look different. For example, in 2x² + 7x + 3, writing 2x² + 6x + x + 3 and grouping as (2x² + 6x) + (x + 3) gives 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1) — the same brackets, just in a different order.
Can I use the ac method for a quadratic where a = 1?
Yes, it still works. With a = 1, ac = c, and you are looking for a factor pair of c that sums to b — exactly the same task as the simpler method. The ac method is simply a generalisation of the basic approach, so you only need to learn one technique.
How does factorising link to finding roots of a quadratic?
If ax² + bx + c = (px + q)(rx + s), then the quadratic equals zero when px + q = 0 or rx + s = 0. So x = −q/p or x = −s/r. Factorising is therefore the fastest method for finding integer or simple-fraction roots, and is typically the first method to try on a GCSE Higher paper.
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