Solving Quadratic Equations by Factorising — AQA GCSE Maths Revision Notes

What you'll learn

This topic covers solving quadratic equations by writing them as a product of two brackets. By the end of this guide you should be able to rearrange an equation to equal zero, factorise it, and read off both solutions.

You should also be able to factorise when the x² coefficient is greater than 1, recognise the difference of two squares, handle equations with no constant term, and know when factorising will not work and another method is needed.

The organising idea is that factorising works because of one fact about zero: if two things multiply to give zero, at least one of them must be zero. Nothing else in mathematics has that property — two numbers multiplying to give 12 could be almost anything, but two multiplying to give 0 force one of them to be 0. That is why the equation must be rearranged so one side is zero before factorising, and why a factorised equation hands you its solutions immediately. Students who set a quadratic equal to 12 and factorise are not doing a slightly worse version of the method; they are doing something that cannot work.

Key terms and definitions

Quadratic equation — an equation whose highest power of x is 2.

Standard form — written as ax² + bx + c = 0, with everything on one side.

Factorise — write an expression as a product of brackets.

Root (or solution) — a value of x satisfying the equation.

Coefficient — the number in front of a term.

Difference of two squares — the pattern a² − b², which factorises to (a + b)(a − b).

Repeated root — a solution occurring twice, when both brackets are identical.

Core concepts

Set the equation to zero first

Rearrange so that one side is zero and the other is in the order ax² + bx + c.

So x² + 5x = −6 becomes x² + 5x + 6 = 0, and 2x² = 7x − 3 becomes 2x² − 7x + 3 = 0.

This step is not a tidying preference; the whole method depends on it. Factorising x² + 5x = −6 to x(x + 5) = −6 tells you nothing, because two numbers multiplying to −6 could be countless pairs.

Factorising when the x² coefficient is 1

Find two numbers that multiply to give c and add to give b.

For x² + 7x + 12: two numbers multiplying to 12 and adding to 7 are 3 and 4, so the factorised form is (x + 3)(x + 4).

The signs follow from the two targets. If c is positive and b is negative, both numbers are negative: x² − 7x + 12 factorises to (x − 3)(x − 4). If c is negative, the two numbers have opposite signs, and the larger takes the sign of b — so x² + 2x − 15 gives (x + 5)(x − 3).

Listing the factor pairs of c systematically is the reliable way to find them, rather than guessing.

Reading off the solutions

Once factorised, set each bracket equal to zero and solve.

For (x + 3)(x + 4) = 0: either x + 3 = 0, giving x = −3, or x + 4 = 0, giving x = −4.

The solutions are x = −3 and x = −4.

Note the sign reversal: a bracket of (x + 3) gives a root of −3, because that is the value making the bracket vanish. Reading the roots straight off as 3 and 4 is the commonest error in the topic.

Checking is quick — substituting x = −3 into the original gives 9 − 21 + 12 = 0. ✓

When the x² coefficient is greater than 1

The two-number method needs adjusting: find two numbers multiplying to a × c and adding to b, then split the middle term and factorise in pairs.

For 2x² + 7x + 3: here a × c = 6, and two numbers multiplying to 6 and adding to 7 are 6 and 1.

Split the middle term: 2x² + 6x + x + 3.

Factorise in pairs: 2x(x + 3) + 1(x + 3).

The bracket (x + 3) is now common, so take it out: (x + 3)(2x + 1).

Setting each bracket to zero gives x = −3 and x = −½.

The pairing step is where care is needed: the two brackets produced must be identical, and if they are not, the split was made with the wrong pair or in the wrong order.

Difference of two squares

When there is no x term and the constant is subtracted, the expression factorises into a matching pair.

x² − 25 = 0 factorises to (x + 5)(x − 5) = 0, giving x = 5 and x = −5.

The general pattern is a² − b² = (a + b)(a − b).

It appears in disguised forms too: 4x² − 9 = 0 factorises to (2x + 3)(2x − 3) = 0, giving x = −1.5 and x = 1.5.

Note the pattern needs a subtraction. There is no equivalent factorisation for x² + 25, which is why that equation has no real solutions.

Equations with no constant term

When c is zero, take out the common factor of x rather than looking for two numbers.

For x² + 6x = 0: factorising gives x(x + 6) = 0, so x = 0 or x = −6.

The solution x = 0 is real and must be included. Dividing both sides by x to get x + 6 = 0 loses it entirely, which is why dividing an equation by a variable is never safe — you may be dividing by zero.

Repeated roots

Sometimes both brackets are the same, and the equation has only one distinct solution.

x² − 6x + 9 = 0 factorises to (x − 3)(x − 3) = 0, so x = 3 is the only root, occurring twice.

Graphically this is the case where the parabola touches the x-axis rather than crossing it, with its turning point sitting exactly on the axis.

When factorising will not work

Factorising only succeeds when the roots are rational. Many quadratics have irrational roots and simply will not factorise, however long you search.

If no pair of whole numbers gives the right product and sum, stop looking and use the quadratic formula or completing the square instead.

The question usually signals which method it wants: "solve by factorising" is explicit; "give your answer to 2 decimal places" or "in surd form" points to the formula or completing the square, since factorising produces exact rational answers that need no rounding.

Worked examples

Example 1: Rearranging first

Solve x² + 3x = 10.

Rearrange so one side is zero: x² + 3x − 10 = 0.

Two numbers multiplying to −10 and adding to 3 are 5 and −2, since the negative product means opposite signs.

So (x + 5)(x − 2) = 0.

Setting each bracket to zero gives x = −5 and x = 2.

Check the second in the original: 4 + 6 = 10. ✓ Factorising the unrearranged version would have been impossible.

Example 2: A coefficient in front of x²

Solve 3x² − 10x + 8 = 0.

Here a × c = 3 × 8 = 24, and two numbers multiplying to 24 and adding to −10 are −6 and −4.

Split the middle term: 3x² − 6x − 4x + 8.

Factorise in pairs: 3x(x − 2) − 4(x − 2).

The common bracket is (x − 2), so the factorised form is (x − 2)(3x − 4).

Setting each to zero gives x = 2 and 3x = 4, so x = 2 and x = 4/3.

Note the second bracket gives a fractional root, which is normal when a is greater than 1.

Example 3: No constant term

Solve 2x² = 8x.

Rearrange to zero: 2x² − 8x = 0.

Take out the common factor: 2x(x − 4) = 0.

Setting each factor to zero gives 2x = 0, so x = 0, and x − 4 = 0, so x = 4.

The solutions are x = 0 and x = 4.

Dividing the original by 2x would have given x = 4 alone, losing the root at zero — a mistake worth a mark and easily avoided by never dividing by a variable.

Common mistakes and how to avoid them

Factorising before setting the equation to zero. The method depends on the zero.

Reading the roots without reversing the sign. (x + 3) gives a root of −3.

Dividing by x and losing a solution. Factorise instead; x = 0 may be a genuine root.

Giving only one solution. A quadratic usually has two; check both brackets.

Getting the signs wrong in the two-number search. A negative constant means opposite signs.

Mismatched brackets when splitting the middle term. The two brackets must be identical before factoring out.

Persisting with factorising when it will not work. Switch to the formula or completing the square.

Exam technique for "Solving Quadratics by Factorising"

Write the rearranged equation in standard form as your first line, even when it is already close. It is often worth a method mark and it makes the two-number search possible.

State the two numbers you are looking for — "multiply to 12, add to 7" — so the examiner can follow the reasoning.

Write both brackets set to zero as separate lines rather than jumping to the answers, since that is where the sign reversal happens.

Check at least one solution by substituting into the original equation.

Give both solutions, and present them clearly as x = something or x = something.

If no factor pair works within a reasonable time, switch methods rather than continuing to search.

Quick revision summary

Factorising works because if two things multiply to zero, one of them must be zero — which is why the equation must be rearranged to equal zero first.

With an x² coefficient of 1, find two numbers that multiply to c and add to b. A negative c means the two numbers have opposite signs.

Set each bracket to zero and solve, remembering the sign reverses: (x + 3) gives a root of −3.

When the x² coefficient is greater than 1, find two numbers multiplying to a × c and adding to b, split the middle term, and factorise in pairs — the two brackets produced must match.

Difference of two squares: x² − 25 = (x + 5)(x − 5). It needs a subtraction, so x² + 25 does not factorise.

With no constant term, take out the common x and keep the root at x = 0. Never divide an equation by a variable.

If no factor pair works, the roots are not rational — use the quadratic formula or completing the square.