Factorising and Solving Quadratic Equations: GCSE Maths

A quadratic equation has the form ax² + bx + c = 0. Factorising means writing it as a product of two brackets — then each bracket can equal zero to find the solutions. For most GCSE questions where a = 1, you are looking for two numbers that multiply to c and add to b.

What makes an equation quadratic?

An equation is quadratic when the highest power of the variable is 2. The standard form is ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0.

Equation type | Highest power | Example
Linear | 1 | 3x + 7 = 0
Quadratic | 2 | x² − 5x + 6 = 0
Cubic | 3 | x³ + 2x² − x = 0

Before factorising, always rearrange so that everything is on one side and the equation equals zero. For example, x² + 3x = 10 must first be written as x² + 3x − 10 = 0.

How do you factorise x² + bx + c?

When the coefficient of x² is 1 (i.e. a = 1), you are looking for two numbers p and q such that:

p × q = c (they multiply to the constant term), and

p + q = b (they add to the coefficient of x).

Once you have found p and q, the factorised form is (x + p)(x + q).

Worked example — factorise x² + 7x + 12:

You need two numbers that multiply to 12 and add to 7.

List factor pairs of 12: (1, 12), (2, 6), (3, 4). Check the sums: 1 + 12 = 13, 2 + 6 = 8, 3 + 4 = 7 ✓

So p = 3 and q = 4.

Factorised form: (x + 3)(x + 4)

Check by expanding: (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 ✓

A note on signs: When c is positive, p and q have the same sign (both positive if b > 0, both negative if b < 0). When c is negative, p and q have opposite signs. Keeping track of signs is where most errors occur — work through every factor pair systematically rather than guessing.

How do you solve the factorised equation?

Once you have written the quadratic as a product of two brackets, you use the zero-product property: if A × B = 0, then either A = 0 or B = 0 (or both).

Worked example — solve x² + 7x + 12 = 0:

Factorise: (x + 3)(x + 4) = 0

Set each bracket equal to zero:

x + 3 = 0 → x = −3

x + 4 = 0 → x = −4

The two solutions are x = −3 and x = −4. These are called the roots of the equation.

You can verify each root by substituting back: (−3)² + 7(−3) + 12 = 9 − 21 + 12 = 0 ✓ and (−4)² + 7(−4) + 12 = 16 − 28 + 12 = 0 ✓

How do you factorise ax² + bx + c when a ≠ 1?

When the coefficient of x² is not 1, use the ac method (also called splitting the middle term).

Worked example — factorise 2x² + 7x + 3:

Multiply a × c: 2 × 3 = 6.

Find two numbers that multiply to 6 and add to 7 (the coefficient of x): 1 and 6 (1 × 6 = 6, 1 + 6 = 7 ✓).

Rewrite the middle term using these two numbers: 2x² + x + 6x + 3.

Factorise by grouping in pairs:

x(2x + 1) + 3(2x + 1)

= (x + 3)(2x + 1)

Check by expanding: (x + 3)(2x + 1) = 2x² + x + 6x + 3 = 2x² + 7x + 3 ✓

Solving: Set each bracket to zero: x + 3 = 0 → x = −3, and 2x + 1 = 0 → x = −½.

The solutions are x = −3 or x = −½.

What is the difference of two squares?

The difference of two squares is a special factorisation pattern worth recognising immediately:

a² − b² = (a + b)(a − b)

It applies whenever an expression has two perfect-square terms separated by a minus sign — and no middle term.

Expression | Factorised form | Check
x² − 9 | (x + 3)(x − 3) | (x+3)(x−3) = x² − 9 ✓
x² − 25 | (x + 5)(x − 5) | (x+5)(x−5) = x² − 25 ✓
4x² − 49 | (2x + 7)(2x − 7) | (2x)² − 7² = 4x² − 49 ✓
9x² − 1 | (3x + 1)(3x − 1) | (3x)² − 1² = 9x² − 1 ✓

To spot this pattern, ask two questions: Are both terms perfect squares? Is the operation between them subtraction? If yes to both, you have a difference of two squares.

Worked example — solve x² − 25 = 0:

Recognise the difference of two squares: x² − 25 = (x + 5)(x − 5)

Set each bracket to zero: x + 5 = 0 → x = −5, and x − 5 = 0 → x = 5

Solutions: x = −5 or x = 5

When should you use the quadratic formula instead?

Factorising is elegant and quick when the roots are integers or simple fractions. However, not every quadratic factorises neatly — the roots may be surds or decimals. In those cases, use the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The expression under the square root, b² − 4ac, is called the discriminant:

Discriminant | What it means
b² − 4ac > 0 | Two distinct real solutions
b² − 4ac = 0 | One repeated solution (a "double root")
b² − 4ac < 0 | No real solutions

A useful strategy: calculate the discriminant first. If it is a perfect square (e.g. 4, 9, 25), factorising will work. If it is not a perfect square, reach for the formula. The quadratic formula always works — factorising is a shortcut that sometimes fails.

Frequently asked questions

What if the quadratic doesn't factorise neatly?

If you cannot find two integers that multiply to c and add to b after checking all factor pairs, the quadratic does not factorise over the integers. In that case, use the quadratic formula or completing the square to find the solutions. Both methods always produce an answer if real solutions exist. At GCSE, a question that says "give your answer to 2 decimal places" is signalling that the formula is required rather than factorising.

Why do you set each bracket equal to zero?

The zero-product property is the key: if two numbers multiply together to give zero, at least one of them must be zero. No two non-zero numbers can multiply to give zero — this is a fundamental property of real numbers. Once you have (x + 3)(x + 4) = 0, the only way for the product to be zero is if (x + 3) = 0 or (x + 4) = 0. That is why setting each bracket to zero, separately, always gives the correct solutions.

What is the difference of two squares and how do I spot it?

The difference of two squares is the pattern a² − b² = (a + b)(a − b). Spot it by checking: (1) does the expression have exactly two terms? (2) are both terms perfect squares? (3) is the sign between them a minus? If all three are true, you have a difference of two squares. Note that a sum of two squares, a² + b², does not factorise over the real numbers — the minus sign is essential.

Can a quadratic have more than two solutions?

No. A quadratic equation has at most two real solutions. This follows from the fundamental theorem of algebra: a polynomial of degree n has at most n roots. Since a quadratic has degree 2, it can have 0, 1, or 2 real solutions (determined by the discriminant), but never 3 or more. If you find three values that appear to satisfy a quadratic, check your arithmetic — at least one will be wrong.

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