To factorise x² + bx + c, find two integers that multiply to c and add to b, then write (x + p)(x + q). Expanding the answer must give back the original quadratic. This is the standard route to solving quadratic equations and simplifying algebraic fractions.
It builds on taking out a common factor and feeds into simplifying algebraic fractions.
Why does the sum-and-product method work?
Expand (x + p)(x + q) and you get x² + px + qx + pq, which is x² + (p + q)x + pq. So the middle coefficient b is the sum p + q, and the constant c is the product pq.
Factorising is the same expansion read backwards. You are looking for the pair p and q that fits both conditions at once.
How to factorise, step by step
- Take out any common factor first.
- Write b and c from x² + bx + c, including signs.
- List factor pairs of c, including the signs that give the right product.
- Pick the pair that adds to b.
- Write (x + p)(x + q) and expand to check.
Worked example
Factorise x² + 2x − 15
Step 1, common factor: none besides 1.
Step 2, b and c: b = +2 and c = −15.
Step 3, factor pairs with product −15:
| Pair | Sum |
|---|---|
| 1 and −15 | −14 |
| −1 and 15 | 14 |
| 3 and −5 | −2 |
| −3 and 5 | 2 |
Step 4, choose: the pair −3 and 5 adds to 2.
Step 5, answer:
(x − 3)(x + 5)
Check: expand. x² + 5x − 3x − 15 = x² + 2x − 15. It matches the original.
The mistake to watch for
A common slip is to ignore the sign of b and use the pair with the right product but the wrong sum.
Mistaken answer: x² − 7x + 12 = (x + 3)(x + 4)
The student found 3 × 4 = 12 and stopped. But (x + 3)(x + 4) expands to x² + 7x + 12, so the middle term has the wrong sign.
The correction is to check the sum with signs. Since c = +12 is positive, both numbers share a sign, and b = −7 is negative, so both are negative.
The pair −3 and −4 multiplies to 12 and adds to −7. So the answer is (x − 3)(x − 4).
Always expand your answer once. It takes a few seconds and catches this error every time.
A variation with a leading coefficient
When the coefficient of x² is not 1, the method needs a small adjustment.
For 2x² + 7x + 3, multiply a and c to get 6. Find two numbers with product 6 and sum 7 (they are 1 and 6). Then split the middle term: 2x² + x + 6x + 3.
Group as x(2x + 1) + 3(2x + 1), which gives (2x + 1)(x + 3). Expanding checks it: 2x² + 6x + x + 3 = 2x² + 7x + 3.
Check yourself
Try these without a calculator, then open each answer.
1. Factorise x² + 7x + 10
Show answer
Both signs are positive. Pairs with product 10: 1 and 10 (sum 11), 2 and 5 (sum 7).
(x + 2)(x + 5)
2. Factorise x² − 2x − 24
Show answer
c is negative, so the signs differ. b is −2, so the larger number is negative. Pairs: 4 and 6 give a difference of 2, so the pair is −6 and +4.
(x − 6)(x + 4)
Check: x² + 4x − 6x − 24 = x² − 2x − 24.
3. Factorise x² − 49
Show answer
Write it as x² + 0x − 49. The pair −7 and 7 multiplies to −49 and adds to 0.
(x − 7)(x + 7)
This is called a difference of two squares.
Where this leads next
Use the quadratic structure explorer to test your factor pair against the roots and the graph. Then continue to simplifying an algebraic fraction with stated exclusions, where these factors do the work. The non-calculator working trainer is handy for the arithmetic inside factor pairs.
Students often understand this method on a clean example and lose it on a question with awkward signs. That is where a teacher can help in online one-to-one Mathematics tuition.