Introduction to Factorising Revision Notes | GCSE AQA Higher Further Maths

1. What Factorising Actually Does

Expanding and factorising are opposites. Expanding takes $(x+3)(x+2)$ and multiplies it out to give $x^2 + 5x + 6$. Factorising runs that process in reverse — you start with $x^2 + 5x + 6$ and recover the two brackets.

Think of factorising as undoing a multiplication. Every factorised form is a product of simpler pieces, which is exactly what examiners mean when they write the command word factorise. Your answer must always be written as a product — brackets multiplied together, not terms added together.

There are four main methods at GCSE: taking out a common factor, factorising a quadratic with leading coefficient 1, factorising a quadratic with leading coefficient other than 1, and the difference of two squares. Choosing the right method quickly is a skill in itself — and that decision process is what this lesson builds.

2. Method 1 — Taking Out a Common Factor

The first thing to check with any expression is whether all the terms share a common factor. If they do, that factor comes out in front of a bracket. This is the simplest and fastest factorising method, and it often appears as the first step inside harder problems too.

To find the highest common factor (HCF), look at the numbers and the letters separately. For $6x^2 + 9x$, the HCF of 6 and 9 is 3, and both terms contain at least one $x$, so the HCF of the whole expression is $3x$.

Worked example: Factorising by Taking Out the HCF

Factorise $6x^2 + 9x$ fully.

1. HCF of $6x^2$ and $9x$: numbers give HCF$(6,9) = 3$; letters give HCF$(x^2, x) = x$. So HCF $= 3x$. — Always find the largest factor shared by every term — this earns the method mark [M1].
2. $6x^2 + 9x = 3x(\quad)$ — Place the HCF in front. The bracket will hold what remains after dividing each term by $3x$.
3. $6x^2 \div 3x = 2x$ and $9x \div 3x = 3$ — Each division gives one term inside the bracket.
4. $6x^2 + 9x = 3x(2x + 3)$ — Final answer [A1]. Verify by expanding: $3x \times 2x + 3x \times 3 = 6x^2 + 9x$. ✓

Answer: $3x(2x + 3)$

A common slip here is taking out a factor that is not the highest one. For example, factorising $6x^2 + 9x$ as $3(2x^2 + 3x)$ is not fully factorised because $x$ can still come out. Examiners specifically ask for full factorisation, so always double-check that nothing further can be removed from the bracket.

3. Method 2 — Factorising Quadratics ($x^2 + bx + c$)

When the expression is a quadratic with a leading coefficient of 1, the goal is to find two numbers that multiply to give $c$ and add to give $b$. Those two numbers go into the brackets $(x + p)(x + q)$.

Worked example: Factorising a Quadratic with Leading Coefficient 1

Factorise $x^2 - 7x + 12$.

1. $b = -7$, $c = 12$ — We need two numbers that multiply to $+12$ and add to $-7$.
2. $1 \times 12,\; 2 \times 6,\; 3 \times 4,\; (-1) \times (-12),\; (-2) \times (-6),\; (-3) \times (-4)$ — Include negative pairs — the product of two negatives is positive, which matches $c = +12$.
3. $(-3) + (-4) = -7$ ✓ and $(-3) \times (-4) = 12$ ✓ — The pair $-3$ and $-4$ satisfies both conditions [M1].
4. $x^2 - 7x + 12 = (x - 3)(x - 4)$ — Final answer [A1]. Expand to verify: $(x-3)(x-4) = x^2 - 4x - 3x + 12 = x^2 - 7x + 12$. ✓

Answer: $(x - 3)(x - 4)$

Notice how the sign of $c$ tells you something immediately. If $c$ is positive, both bracket numbers share the same sign (both positive or both negative). If $c$ is negative, the bracket numbers have opposite signs. Spotting this before listing pairs saves time in an exam.

4. Method 3 — Factorising Quadratics ($ax^2 + bx + c$, $a \neq 1$)

When the coefficient of $x^2$ is not 1, the factor-pair approach needs a small adjustment. The most reliable method at GCSE is the product-and-sum method (sometimes called the AC method): multiply $a$ and $c$ together, then find two numbers that multiply to give $ac$ and add to give $b$.

Worked example: Factorising a Quadratic where $a \neq 1$

Factorise $2x^2 + 7x + 3$.

1. $a = 2$, $c = 3$, so $ac = 2 \times 3 = 6$ — We now need two numbers that multiply to $6$ and add to $b = 7$.
2. Factor pairs of $6$: $(1, 6)$ — sum $= 7$ ✓ — The pair $1$ and $6$ works [M1].
3. $2x^2 + 7x + 3 = 2x^2 + x + 6x + 3$ — Replace $7x$ with $x + 6x$. The order can be swapped — both work.
4. $= x(2x + 1) + 3(2x + 1)$ — Group the first two terms and the last two terms, then take out the common factor from each pair [M1].
5. $= (2x + 1)(x + 3)$ — Final answer [A1]. Expand to verify: $(2x+1)(x+3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3$. ✓

Answer: $(2x + 1)(x + 3)$

5. Method 4 — Difference of Two Squares

There is a special pattern worth memorising because it appears frequently and can be factorised in one step. When you see two perfect square terms with a subtraction between them and nothing else, you have a difference of two squares.

Worked example: Applying the Difference of Two Squares

Factorise $9x^2 - 25$.

1. $9x^2 = (3x)^2$ and $25 = 5^2$ — Both terms are perfect squares and they are separated by a minus sign — the pattern applies [M1].
2. $a = 3x$, $b = 5$ — $a$ is the square root of the first term; $b$ is the square root of the second.
3. $9x^2 - 25 = (3x + 5)(3x - 5)$ — Final answer [A1]. Expand to verify: $(3x+5)(3x-5) = 9x^2 - 15x + 15x - 25 = 9x^2 - 25$. ✓

Answer: $(3x + 5)(3x - 5)$

The difference of two squares also applies to purely numerical expressions. For instance, $100 - 49 = (10+7)(10-7) = 17 \times 3 = 51$, which is a neat mental arithmetic trick. At GCSE, you are more likely to see it with algebraic terms, but recognising the pattern in any form is the key skill.

6. Choosing the Right Method

In an exam, you will not be told which method to use. The expression itself gives you the clues. Work through this decision sequence every time.

• First: check whether all terms share a common factor. If yes, take it out. This may be the only step needed, or it may simplify what remains.
• Then: count the terms. Two terms with a minus sign and both perfect squares? Use difference of two squares.
• Three terms with leading coefficient 1? Use the factor-pair method ($p \times q = c$, $p + q = b$).
• Three terms with leading coefficient other than 1? Use the product-and-sum (AC) method.

7. Deliberate Practice

Now that all four methods are in place, the goal is to practise selecting and applying them fluently. Work through each question below, deciding on the method before writing anything down.

8. Exam Technique and Common Mistakes

Factorising questions are reliable mark-earners when the method is applied carefully. The mistakes below account for the majority of dropped marks — recognise them now so they do not cost you in the exam.

Factorising is the foundation for solving quadratic equations, simplifying algebraic fractions, and sketching curves — all of which build directly on the four methods covered here. Every time you factorise an expression correctly, you are applying the same core skill: rewriting a sum as a product.