Expanding and Factorising

Expanding brackets and factorising are reverse operations — two of the most important algebraic skills in GCSE Mathematics. This lesson covers single brackets, double brackets, the difference of two squares, and factorising quadratics including those where $a \neq 1 ($a=1(Higher tier).

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Key Vocabulary

Term	Meaning
Expand	Multiply out the brackets
Factorise	Write as a product of factors (put back into brackets)
Common factor	A factor shared by every term
Quadratic expression	An expression of the form $ax^{2} + bx + c$ax2+bx+c
Difference of two squares	An expression of the form $a^{2} - b^{2}$a2−b2

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Expanding Single Brackets

Multiply the term outside the bracket by every term inside.

Worked Example 1

Expand: 3(2x + 5)

$= 3 \times 2x + 3 \times 5 =$=3×2x+3×5= 6x + 15

Worked Example 2

Expand: −2x(3x − 4y + 1)

= (−2x)(3x) + (−2x)(−4y) + (−2x)(1)

$= -6x^{2} + 8xy - 2x$=−6x2+8xy−2x

Answer: $-6x^{2} + 8xy - 2x$−6x2+8xy−2x

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Expanding Double Brackets

Multiply each term in the first bracket by each term in the second. A useful method is FOIL (First, Outer, Inner, Last).

Worked Example 3

Expand: (x + 3)(x + 5)

$= x^{2} + 5x + 3x + 15 =$=x2+5x+3x+15= $x^{2} + 8x + 15$x2+8x+15

Worked Example 4

Expand: (2x − 1)(3x + 4)

$= 6x^{2} + 8x - 3x - 4 =$=6x2+8x−3x−4= $6x^{2} + 5x - 4$6x2+5x−4

Worked Example 5

Expand: $(x - 7)^{2}$(x−7)2

$(x - 7)(x - 7) = x^{2} - 7x - 7x + 49 =$(x−7)(x−7)=x2−7x−7x+49= $x^{2} - 14x + 49$x2−14x+49

Common mistake: $(x - 7)^{2} \neq x^{2} - 49$(x−7)2=x2−49. You must multiply out the double bracket — don't just square each term.

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Expanding Triple Brackets [H]

Expand two brackets first, then multiply the result by the third.

Worked Example 6

Expand: (x + 1)(x + 2)(x − 3)

Step 1: $(x + 1)(x + 2) = x^{2} + 3x + 2$(x+1)(x+2)=x2+3x+2

Step 2: $(x^{2} + 3x + 2)(x - 3)$(x2+3x+2)(x−3)

$= x^{3} - 3x^{2} + 3x^{2} - 9x + 2x - 6$=x3−3x2+3x2−9x+2x−6

= $x^{3} - 7x - 6$x3−7x−6

Let's verify: the $x^{2}$x2 terms cancel $(-3x^{2} + 3x^{2} = 0)$(−3x2+3x2=0), the x terms are −9x + 2x = −7x. Correct.

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The Difference of Two Squares

$a^{2} - b^{2} = (a + b)(a - b)$a2−b2=(a+b)(a−b)

This works both ways — for expanding and for factorising.

Worked Example 7

Expand: $(x + 6)(x - 6) = x^{2} - 36$(x+6)(x−6)=x2−36

Worked Example 8

Factorise: $9x^{2} - 25$9x2−25

$= (3x)^{2} - 5^{2} =$=(3x)2−52= (3x + 5)(3x − 5)

Worked Example 9

Factorise: $4y^{2} - 49z^{2}$4y2−49z2

$= (2y)^{2} - (7z)^{2} =$=(2y)2−(7z)2= (2y + 7z)(2y − 7z)

Factorising — Decision Flow

flowchart TD
    EXPR[Expression to factorise] --> HCF{"Is there a common factor<br/>across all terms?"}
    HCF -->|Yes| TAKE["Take out HCF first<br/>e.g. 2x sq minus 18 becomes 2 times x sq minus 9"]
    HCF -->|No| TERMS{How many terms?}
    TAKE --> TERMS
    TERMS -->|Two terms| DOSQ{"Difference of two squares<br/>a sq minus b sq?"}
    DOSQ -->|Yes| APPLY_DOS[Apply a sq minus b sq = a + b times a minus b]
    DOSQ -->|No| DONE1["Already factorised<br/>or cannot factorise further"]
    TERMS -->|Three terms| QUAD{Is it ax sq + bx + c?}
    QUAD -->|a = 1| SIMPLE["Find two numbers<br/>multiplying to c, adding to b"]
    QUAD -->|a not equal 1 H| GROUP["Use GROUPING method<br/>find numbers multiplying to ac, adding to b<br/>then split middle term"]
    TERMS -->|Four terms| FACGRP[Factorise by GROUPING in pairs]
    APPLY_DOS --> CHECK[Verify by expanding back]
    SIMPLE --> CHECK
    GROUP --> CHECK
    FACGRP --> CHECK

The Edexcel command word "factorise fully" is the trigger to look for the HCF first, then continue factorising the remaining bracket as far as possible.

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Factorising — Taking Out a Common Factor

Look for the highest common factor (HCF) of all terms.

Worked Example 10

Factorise: $6x^{2} + 15x$6x2+15x

HCF = 3x

= 3x(2x + 5)

Worked Example 11

Factorise: $12a^{3}b^{2} - 8a^{2}b + 4ab$12a3b2−8a2b+4ab

HCF = 4ab

= $4ab(3a^{2}b - 2a + 1)$4ab(3a2b−2a+1)

Check by expanding back: $4ab \times 3a^{2}b = 12a^{3}b^{2}$4ab×3a2b=12a3b2 ✓; $4ab \times (-2a) = -8a^{2}b$4ab×(−2a)=−8a2b ✓; $4ab \times 1 = 4ab$4ab×1=4ab ✓

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Factorising Quadratics: $x^{2} + bx + c$x2+bx+c

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

Worked Example 12

Factorise: $x^{2} + 7x + 12$x2+7x+12

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

Factors of 12: $1 \times 12$1×12, $2 \times 6$2×6, $3 \times 4$3×4 ← these add to 7 ✓

Answer: (x + 3)(x + 4)

Worked Example 13

Factorise: $x^{2} - 3x - 10$x2−3x−10

We need two numbers that multiply to −10 and add to −3.

Options: $(-5) \times 2 = -10$(−5)×2=−10 and (−5) + 2 = −3 ✓

Answer: (x − 5)(x + 2)

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Factorising Quadratics: $ax^{2} + bx + c (a \neq 1) [H]$ax2+bx+c(a=1)[H]

When the coefficient of $x^{2}$x2 is not 1, use the grouping method (splitting the middle term).

Method

1. Multiply $a \times c$a×c.
2. Find two numbers that multiply to ac and add to b.
3. Rewrite the middle term using those two numbers.
4. Factorise by grouping.

Worked Example 14

Factorise: $6x^{2} + 11x + 3$6x2+11x+3

Step 1: $a \times c = 6 \times 3 = 18$a×c=6×3=18

Step 2: Two numbers that multiply to 18 and add to $11 \to$11→ 9 and 2

Step 3: $6x^{2} + 9x + 2x + 3$6x2+9x+2x+3

Step 4: 3x(2x + 3) + 1(2x + 3) = (3x + 1)(2x + 3)

Check: $3x \times 2x = 6x^{2}$3x×2x=6x2; $3x \times 3 + 1 \times 2x = 9x + 2x = 11x$3x×3+1×2x=9x+2x=11x; $1 \times 3 = 3$1×3=3 ✓

Worked Example 15

Factorise: $4x^{2} - 12x + 5$4x2−12x+5

Step 1: $a \times c = 4 \times 5 = 20$a×c=4×5=20

Step 2: Two numbers that multiply to 20 and add to $-12 \to$−12→ −10 and −2

Step 3: $4x^{2} - 10x - 2x + 5$4x2−10x−2x+5

Step 4: 2x(2x − 5) − 1(2x − 5) = (2x − 1)(2x − 5)

Check: $2x \times 2x = 4x^{2}$2x×2x=4x2; 2x(−5) + (−1)(2x) = −10x − 2x = −12x; (−1)(−5) = 5 ✓

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Factorising by Grouping (Four Terms)

When an expression has four terms, try grouping in pairs.

Worked Example 16

Factorise: xy + 3x + 2y + 6

= x(y + 3) + 2(y + 3)

= (x + 2)(y + 3)

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Common Mistakes and Misconceptions