Factorising is usually the quickest way to solve a quadratic equation. You write the quadratic as two brackets multiplied together, such as (x + 2)(x − 3) = 0, then use one simple fact to finish.
The zero product rule
If two numbers multiply to give 0, at least one of them must be 0. There is no other way to get 0 from a multiplication.
So if (x + 2)(x − 3) = 0, then either the first bracket is 0 or the second bracket is 0:
- x + 2 = 0, so x = −2
- x − 3 = 0, so x = 3
Check: (−2 + 2)(−2 − 3) = 0 × (−5) = 0 ✓ and (3 + 2)(3 − 3) = 5 × 0 = 0 ✓.
Notice the signs flip: the bracket (x + 2) gives x = −2, and (x − 3) gives x = 3.
Try (x + 2)(x − 3) = 0 in the Quadratic Solver →
Getting the brackets when x² has no number in front
Most questions give the equation expanded, like x² − x − 6 = 0. To factorise x² + bx + c, find two numbers that multiply to c and add to b. Those numbers go in the brackets.
Example 1
Solve x² + 5x + 6 = 0.
Two numbers that multiply to 6 and add to 5: 2 and 3.
So x² + 5x + 6 = (x + 2)(x + 3) = 0, which gives x = −2 or x = −3.
Example 2: negative numbers
Solve x² − x − 6 = 0.
Two numbers that multiply to −6 and add to −1: 2 and −3. One must be negative because the product is negative.
So (x + 2)(x − 3) = 0, which gives x = −2 or x = 3.
A quick way to search: list the factor pairs of c, then look for the pair that adds or subtracts to make b. For c = −6 the pairs are 1 and 6, and 2 and 3, and only 2 and 3 can make −1 (as 2 + (−3)).
Two special cases
No number term: take out x
Solve x² − 5x = 0.
Both terms contain x, so factorise as x(x − 5) = 0. Then x = 0 or x = 5.
Don’t divide both sides by x. That loses the solution x = 0.
Difference of two squares
Solve x² − 9 = 0.
x² − 9 = (x + 3)(x − 3), because the middle terms cancel. So x = −3 or x = 3.
When x² has a number in front
For ax² + bx + c with a not equal to 1, use the a × c method: find two numbers that multiply to a × c and add to b, use them to split the middle term, and factorise in pairs.
Example 3
Solve 2x² − 3x − 2 = 0.
a × c = 2 × (−2) = −4. Two numbers that multiply to −4 and add to −3: −4 and 1.
Split the middle term: 2x² − 4x + x − 2 = 0.
Factorise each pair: 2x(x − 2) + 1(x − 2) = 0, so (2x + 1)(x − 2) = 0.
Set each bracket to zero: 2x + 1 = 0 gives x = −½, and x − 2 = 0 gives x = 2.
See Example 3 worked in the Quadratic Solver →
One repeated solution
x² − 6x + 9 = 0 factorises as (x − 3)(x − 3) = (x − 3)². Both brackets give the same answer, so there is one repeated solution, x = 3.
Make one side zero first
The zero product rule only works when the brackets multiply to 0. For (x + 1)(x + 2) = 6 you can’t set each bracket to 6. Expand and rearrange first: x² + 3x + 2 = 6 becomes x² + 3x − 4 = 0, which factorises as (x + 4)(x − 1) = 0. So x = −4 or x = 1.
When it doesn’t factorise
Many quadratics don’t factorise with whole numbers. For x² − 2x − 1 = 0 there are no two whole numbers that multiply to −1 and add to −2. A quick test: work out b² − 4ac. If it is a square number (0, 1, 4, 9, 16, …) the quadratic factorises. Here it is 4 + 4 = 8, which isn’t square, so use the quadratic formula instead.
The Quadratic Solver shows factorising step by step for any equation you type, including ones already in brackets, and tells you when an equation doesn’t factorise.