Factorising Revision Notes for AQA GCSE Further Maths

GCSE Algebra II: Factorising

What is factorising?

Factorising is the process of writing an algebraic expression as a product of its factors using brackets. Think of it as the reverse of expanding brackets - instead of multiplying out, you're breaking an expression down into simpler parts that multiply together.

This skill builds on your knowledge of finding common factors from earlier topics. When you factorise, you're essentially asking "what multiplies together to give me this expression?"

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Factorising is one of the most important algebraic skills you'll learn. It's used extensively in solving equations, simplifying expressions, and working with functions in higher mathematics.

Basic factorising by grouping

The simplest type of factorising involves expressions with four terms that can be grouped in pairs. Here's how to approach this systematically:

Method:

Group the terms in pairs
Take out the common factor from each pair
Look for a common bracket that appears in both groups
Factor out this common bracket

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Worked Example: Grouping Method

Factorise $xa + xb + ya + yb$

Step 1: Group the terms: $(xa + xb) + (ya + yb)$

Step 2: Take out common factors: $x(a + b) + y(a + b)$

Step 3: Notice that $(a + b)$ appears in both terms, so factor it out: $(a + b)(x + y)$

The rectangular diagram visualisation can help you understand how this process relates to areas of rectangles, showing what's happening when you factorise.

Factorising quadratics by splitting the middle term

This method works for quadratic expressions of the form $x^2 + bx + c$ . The key is learning how to split the middle term effectively.

Method:

Look at the coefficient of $x$ and the constant term
Find two numbers that multiply to give the constant AND add to give the coefficient of $x$
Split the middle term using these two numbers
Group and factorise

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Worked Example: Splitting the Middle Term

Factorise $x^2 + 6x + 8$

Step 1: You need two numbers that multiply to give 8 and add to give 6.

Step 2: The pairs that multiply to give 8 are: $1 \times 8$ , $2 \times 4$

Step 3: Check which pair adds to 6: $1 + 8 = 9$ (no), $2 + 4 = 6$ (yes!)

Step 4: So split $6x$ as $4x + 2x$ : $x^2 + 6x + 8 = x^2 + 4x + 2x + 8$ $= x(x + 4) + 2(x + 4)$ $= (x + 4)(x + 2)$

Why does this work? The numbers 4 and 2 have a sum of 6 (the coefficient of $x$ ) and a product of 8 (the constant term). This is the only combination that satisfies both conditions.

Finding factor pairs for negative constants

When the constant term is negative, one of your factors must be positive and one negative. This requires a slightly different approach.

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Worked Example: Negative Constants

Factorise $x^2 - 7x - 18$

Step 1: Look for pairs of numbers with a product of $-18$ :

$1$ and $-18$
$2$ and $-9$
$3$ and $-6$
$6$ and $-3$
$9$ and $-2$
$18$ and $-1$

Step 2: You need the pair that adds to $-7$ . Check: $2 + (-9) = -7$ ✓

Step 3: So: $x^2 - 7x - 18 = x^2 + 2x - 9x - 18$ $= x(x + 2) - 9(x + 2)$ $= (x + 2)(x - 9)$

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Sign Changes: Notice the sign change - when you have a negative in front of the 9, the final bracket becomes $(x - 9)$ . This is a common source of errors, so always double-check your signs!

The difference of two squares

This is a special pattern that you should learn to recognise immediately. It has the form $a^2 - b^2$ , and it always factorises as $(a + b)(a - b)$ .

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Key Formula to Remember: $a^2 - b^2 = (a + b)(a - b)$ This pattern should be memorised and recognised instantly in exams.

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Worked Example: Basic Difference of Squares

Factorise $x^2 - 16$

Step 1: Recognise that $x^2 - 16 = x^2 - 4^2$

Step 2: This fits the pattern $a^2 - b^2$ where $a = x$ and $b = 4$

Step 3: So $x^2 - 16 = (x + 4)(x - 4)$

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Worked Example: With Coefficients

Factorise $4x^2 - 9y^2$

Step 1: First, rewrite in the form $a^2 - b^2$ : $4x^2 - 9y^2 = (2x)^2 - (3y)^2$

Step 2: Apply the formula: $(2x + 3y)(2x - 3y)$

Factorising fully

Sometimes you need to use multiple techniques in sequence. The instruction "factorise fully" tells you there's likely more than one step involved.

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Worked Example: Multiple Steps

Factorise fully $y^5 - 36y^3$

Step 1: Look for common factors first Both terms contain $y^3$ , so factor it out: $y^3(y^2 - 36)$

Step 2: Look at what's left in the bracket $y^2 - 36$ is a difference of two squares: $y^2 - 6^2$

Step 3: Apply the difference of squares formula $y^2 - 36 = (y + 6)(y - 6)$

Final answer: $y^3(y + 6)(y - 6)$

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Why factorise fully? You want to break the expression down as far as possible into factors that cannot be simplified further. This makes expressions easier to work with in equations and other algebraic manipulations.

Advanced techniques for more complex quadratics

When the coefficient of $x^2$ is not 1, you need to modify your approach slightly.

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Worked Example: Coefficient ≠ 1

Factorise $2x^2 - 11x + 15$

Step 1: Here the sum is $-11$ and you need a product of $2 \times 15 = 30$

Step 2: Look for two numbers that multiply to 30 and add to $-11$ : Since both the sum is negative and the product is positive, both numbers must be negative.

Step 3: Try: $-5$ and $-6$ → $(-5) \times (-6) = 30$ ✓ and $(-5) + (-6) = -11$ ✓

Step 4: Split the middle term: $2x^2 - 11x + 15 = 2x^2 - 5x - 6x + 15$ $= x(2x - 5) - 3(2x - 5)$ $= (x - 3)(2x - 5)$

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Worked Example: Two Variables

Factorise $3x^2 - 10xy - 8y^2$

Step 1: Find two numbers that multiply to give $3 \times (-8) = -24$ and add to give $-10$ . These are $-12$ and $2$ : $(-12) \times 2 = -24$ and $(-12) + 2 = -10$

Step 2: The expression becomes: $3x^2 - 12xy + 2xy - 8y^2$ $= 3x(x - 4y) + 2y(x - 4y)$ $= (3x + 2y)(x - 4y)$

Common exam tips and pitfalls

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Essential Exam Strategies:

Always check your work by expanding your answer back out - you should get the original expression
Look for common factors first before attempting other methods. This often makes the remaining factorisation much easier
With difference of squares, remember both terms must be perfect squares (like $x^2$ , $4$ , $9x^2$ , $25y^2$ )
For negative products, remember that one factor is positive and one is negative. The larger number (ignoring signs) goes with the sign of the middle term
When factorising fully, don't stop after the first step - keep looking for further factorisation opportunities

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Key Points to Remember:

Factorising is the reverse of expanding - you're breaking expressions into factors that multiply together
Always look for common factors first before trying other methods
For quadratics $ax^2 + bx + c$ , find two numbers that multiply to give $ac$ and add to give $b$
The difference of squares pattern $a^2 - b^2 = (a + b)(a - b)$ should be memorised
"Factorise fully" means use multiple steps until no further factorisation is possible
Always check your answer by expanding it back out to verify you get the original expression

Factorising (AQA GCSE Further Maths): Revision Notes

GCSE Algebra II: Factorising

What is factorising?

Basic factorising by grouping

Factorising quadratics by splitting the middle term

Finding factor pairs for negative constants

The difference of two squares

Factorising fully

Advanced techniques for more complex quadratics

Common exam tips and pitfalls

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