Factorising by Extracting Common Factors

Learn how to factorise algebraic expressions by extracting the highest common factor of numbers and variables, avoid the vanishing term trap, and check your answers by expanding.

Before you start

Highest Common Factor (HCF): The greatest number that divides exactly into two or more numbers without a remainder. For example, the HCF of $12$ and $18$ is $6$ .
The Distributive Law: $a(b + c) = ab + ac$ . Factorisation is the exact reverse process of expanding brackets.
Dividing a term by a number and a letter: $15xy \div 5x = 3y$ , because $5x \times 3y = 15xy$ . A term divided by itself is $1$ : $7m \div 7m = 1$ .
Checking by expansion: You can always verify your factorisation by multiplying the terms back out. The result must match the original expression exactly. Expanding shows the factorisation is correct; to check it is complete, also make sure the terms inside the bracket have no common factor left.

1. What Is Factorisation?

In algebra, factorisation is the process of writing an expression as a product of its factors. It is the exact reverse of expanding brackets.

When you expand, you remove brackets by multiplying:

$\text{Expanding: } 3(2x + 5) = 6x + 15$

When you factorise, you insert brackets by extracting common factors:

$\text{Factorising: } 6x + 15 = 3(2x + 5)$

Factorising expressions is a crucial algebraic skill used throughout Secondary mathematics to simplify expressions, cancel fractions, and solve equations.

The Process of Extracting Common Factors

To factorise an expression by extracting common factors:

1. Find the Highest Common Factor (HCF) of all the numbers in the terms (the coefficients and any constant term).

2. Find any letter that appears in every term.

3. Multiply the numerical HCF and the common letters together to form the complete common factor.

4. Write the common factor outside a bracket, and divide each original term by this common factor to determine the terms remaining inside.

2. Extracting a Numerical Common Factor

When all terms in an expression share a common whole number factor, that number can be taken out.

For example, in $6x + 9$ :

- The numbers are $6$ and $9$ . Their highest common factor is $3$ .

- Divide each term by $3$ : $6x \div 3 = 2x$ , and $9 \div 3 = 3$ .

- The factorised expression is $3(2x + 3)$ .

Worked Example 1: Extracting Numerical Factors

Factorise each of the following expressions completely:
(a) $8x - 20$
(b) $12a - 18b + 24$
(c) $-14m - 21n$

Solution:

(a) The HCF of $8$ and $20$ is $4$ :

Divide each term by $4$ : $8x \div 4 = 2x$ , and $-20 \div 4 = -5$ :

$8x - 20 = 4(2x - 5)$

(b) The HCF of $12, 18,$ and $24$ is $6$ :

Divide each term by $6$ :

$12a \div 6 = 2a, \quad -18b \div 6 = -3b, \quad 24 \div 6 = 4$

$12a - 18b + 24 = 6(2a - 3b + 4)$

(c) Both terms are negative, and the HCF of $14$ and $21$ is $7$ . Extract $-7$ :

Divide each term by $-7$ :

$-14m \div (-7) = 2m, \quad -21n \div (-7) = 3n$

$-14m - 21n = -7(2m + 3n)$

(a) $4(2x - 5)$ (b) $6(2a - 3b + 4)$ (c) $-7(2m + 3n)$

3. Common Factors with Letters

Terms often share a letter as well as a number. When every term contains the same letter, that letter is part of the common factor.

Expressions such as $6xy + 9x$ are not linear (Lesson 8.1), because $6xy$ has two letters multiplied together, but a common letter can still be taken out of them in the same way.

Consider $6xy + 9x$ :

- Numerical coefficients: $\text{HCF}(6, 9) = 3$ .

- Letters: $6xy = 6 \times x \times y$ , and $9x = 9 \times x$ . The letter $x$ appears in both terms; $y$ does not.

- The complete common factor is $3x$ .

- Divide each term by $3x$ : $6xy \div 3x = 2y$ , and $9x \div 3x = 3$ .

- Result: $6xy + 9x = 3x(2y + 3)$ .

Complete vs. Incomplete Factorisation

An expression is completely factorised only when the terms left inside the bracket have no common factors left (other than $1$ ).

For example, factorising $14mn - 21mp$ :

- If you only extract $7$ : $7(2mn - 3mp)$ . This is incomplete, because both terms inside still share the common factor $m$ .

- The complete common factor is $7m$ :

$14mn - 21mp = 7m(2n - 3p)$

Inside the bracket, $2n$ and $-3p$ have no common factors left. The factorisation is now complete.

Worked Example 2: Extracting Factors Made of a Number and a Letter

Factorise each of the following expressions completely:
(a) $15xy - 20x$
(b) $12ab + 18ac$
(c) $12ab - 8ac + 20ad$

Solution:

(a) Both terms contain $x$ . The HCF of $15$ and $20$ is $5$ , so the common factor is $5x$ :

Divide each term by $5x$ : $15xy \div 5x = 3y$ , and $-20x \div 5x = -4$ :

$15xy - 20x = 5x(3y - 4)$

(b) Both terms contain $a$ . The HCF of $12$ and $18$ is $6$ , so the common factor is $6a$ :

Divide each term by $6a$ : $12ab \div 6a = 2b$ , and $18ac \div 6a = 3c$ :

$12ab + 18ac = 6a(2b + 3c)$

(c) All three terms contain $a$ ; no other letter is in every term. The HCF of $12, 8,$ and $20$ is $4$ , so the common factor is $4a$ :

Divide each term by $4a$ :

$12ab \div 4a = 3b, \quad -8ac \div 4a = -2c, \quad 20ad \div 4a = 5d$

$12ab - 8ac + 20ad = 4a(3b - 2c + 5d)$

Check by expanding: $4a \times 3b - 4a \times 2c + 4a \times 5d = 12ab - 8ac + 20ad$ .

(a) $5x(3y - 4)$ (b) $6a(2b + 3c)$ (c) $4a(3b - 2c + 5d)$

4. The Vanishing Term Trap and Negative Factors

One of the most frequent traps in algebraic factorisation occurs when an entire term is identical to the common factor being extracted.

The Vanishing Term Trap: Anything Divided by Itself is 1

Consider factorising $11a + 22ab - 44ac$ :

- The common factor of all three terms is $11a$ .

- What happens to the first term, $11a$ ?

- Many students mistakenly think $11a$ disappears or becomes $0$ .

- In reality: $11a \div 11a = 1$ . The term becomes $1$ !

$11a + 22ab - 44ac = 11a(1 + 2b - 4c)$

Check by expanding: $11a \times 1 = 11a$ . If you left out the $1$ , expanding $11a(2b - 4c)$ would give $22ab - 44ac$ , losing the original first term entirely!

The same happens with a number: in $5x - 5$ , the HCF is $5$ and $-5 \div 5 = -1$ , so $5x - 5 = 5(x - 1)$ .

Extracting a Negative Common Factor

When the first term is negative, it is often neater to take out a negative common factor.

Dividing by a negative number reverses the sign of every term inside the bracket:

$-30 - 35t = -5(6 + 7t)$

$-4x + 6y = -2(2x - 3y)$

Both $-2(2x - 3y)$ and $2(3y - 2x)$ are correct and complete; a question that wants a negative factor will say so.

Worked Example 3: The Vanishing Term and Negative Factors

Factorise each of the following expressions completely:
(a) $7x + 14xy$
(b) $12pq - 4p$
(c) $6m - 6$
(d) $-18a - 27b - 9$

Solution:

(a) Both terms contain $x$ , and the HCF of $7$ and $14$ is $7$ . The common factor is $7x$ :

Divide: $7x \div 7x = 1$ , and $14xy \div 7x = 2y$ :

$7x + 14xy = 7x(1 + 2y)$

(b) Both terms contain $p$ , and the HCF of $12$ and $4$ is $4$ . The common factor is $4p$ :

Divide: $12pq \div 4p = 3q$ , and $-4p \div 4p = -1$ :

$12pq - 4p = 4p(3q - 1)$

(c) Both coefficients are $6$ , so the common factor is $6$ . Divide: $6m \div 6 = m$ , and $-6 \div 6 = -1$ :

$6m - 6 = 6(m - 1)$

(d) All three terms are negative, and the numbers $18$ , $27$ and $9$ are all divisible by $9$ . Extract $-9$ :

Divide each term by $-9$ :

$-18a \div (-9) = 2a, \quad -27b \div (-9) = 3b, \quad -9 \div (-9) = 1$

$-18a - 27b - 9 = -9(2a + 3b + 1)$

(a) $7x(1 + 2y)$ (b) $4p(3q - 1)$ (c) $6(m - 1)$ (d) $-9(2a + 3b + 1)$

Common Mistakes to Avoid

Incomplete factorisation: Writing $12x + 18 = 2(6x + 9)$ instead of $6(2x + 3)$ . Always check that the terms inside the bracket share no further common factors.
The vanishing term error: Writing $5x + 10xy = 5x(2y)$ by omitting the $1$ . Remember $5x \div 5x = 1$ , so the correct answer is $5x(1 + 2y)$ .
Sign mistakes when extracting a negative factor: Writing $-6a - 9b = -3(2a - 3b)$ . Dividing $-9b$ by $-3$ gives $+3b$ , so the correct form is $-3(2a + 3b)$ .
Forgetting to check by expanding: After factorising, quickly expand your bracket in your head or on scratch paper. If the expansion does not reproduce the original expression, re-check your division. Then make sure the terms inside the bracket have no common factor left, or the factorisation is not complete.
Omitting brackets around multi-term expressions before units: Writing $2x + 3\text{ cm}$ instead of $(2x + 3)\text{ cm}$ .

Practice Questions

Work out each problem on paper before checking your solution. Questions 9 to 12 apply factorisation to a banner, a picture frame, shelves of shuttlecocks and a hall floor.

Question 1: Extracting Numerical Common Factors

Factorise each of the following expressions completely:
(a) $10x + 25$
(b) $21a - 28b$
(c) $16m - 24n + 40$

ShowHide solution

(a) $\text{HCF}(10, 25) = 5$ :
$10x + 25 = 5(2x + 5)$

(b) $\text{HCF}(21, 28) = 7$ :
$21a - 28b = 7(3a - 4b)$

(c) $\text{HCF}(16, 24, 40) = 8$ :
$16m - 24n + 40 = 8(2m - 3n + 5)$

(a) $5(2x + 5)$ (b) $7(3a - 4b)$ (c) $8(2m - 3n + 5)$

Question 2: Extracting a Common Letter

Factorise each expression completely by extracting a common letter:
(a) $xy + 5x$
(b) $3pq - 7q$
(c) $4ab + 9ac$

ShowHide solution

(a) Both terms contain $x$ ; $1$ is the only common number:
$xy + 5x = x(y + 5)$

(b) Both terms contain $q$ ; $\text{HCF}(3, 7) = 1$ :
$3pq - 7q = q(3p - 7)$

(c) Both terms contain $a$ ; $\text{HCF}(4, 9) = 1$ :
$4ab + 9ac = a(4b + 9c)$

(a) $x(y + 5)$ (b) $q(3p - 7)$ (c) $a(4b + 9c)$

Question 3: Common Factors Made of a Number and a Letter

Factorise each expression completely:
(a) $8xy + 12x$
(b) $15ab - 25ac$
(c) $18pq - 24pr$

ShowHide solution

(a) $\text{HCF}(8, 12) = 4$ , and both terms contain $x$ . Common factor is $4x$ :
$8xy + 12x = 4x(2y + 3)$

(b) $\text{HCF}(15, 25) = 5$ , and both terms contain $a$ . Common factor is $5a$ :
$15ab - 25ac = 5a(3b - 5c)$

(c) $\text{HCF}(18, 24) = 6$ , and both terms contain $p$ . Common factor is $6p$ :
$18pq - 24pr = 6p(3q - 4r)$

(a) $4x(2y + 3)$ (b) $5a(3b - 5c)$ (c) $6p(3q - 4r)$

Question 4: Complete vs Incomplete Factorisation

Explain why each of the following factorisations is incomplete, and write down the complete factorisation:
(a) $8xy + 16x = 4(2xy + 4x)$
(b) $18m - 12n = 3(6m - 4n)$

ShowHide solution

(a) The terms inside the bracket ( $2xy$ and $4x$ ) still share a common factor of $2x$ . The complete common factor is $4 \times 2x = 8x$ :
$8xy + 16x = 8x(y + 2)$
Check: $8x \times y + 8x \times 2 = 8xy + 16x$ .

(b) The terms inside the bracket ( $6m$ and $-4n$ ) still share a common factor of $2$ . The HCF of $18$ and $12$ is $6$ :
$18m - 12n = 6(3m - 2n)$
Check: $6 \times 3m - 6 \times 2n = 18m - 12n$ .

(a) Common factor $2x$ was left inside; complete: $8x(y + 2)$ (b) Common factor $2$ was left inside; complete: $6(3m - 2n)$

Question 5: The Vanishing Term Trap

Factorise each expression completely:
(a) $9y + 27yz$
(b) $13 + 26a - 39b$
(c) $5xy - 5x$

ShowHide solution

(a) Common factor is $9y$ . Dividing $9y$ by $9y$ gives $1$ , and $27yz \div 9y = 3z$ :
$9y + 27yz = 9y(1 + 3z)$

(b) $\text{HCF}(13, 26, 39) = 13$ . Dividing $13$ by $13$ leaves $1$ :
$13 + 26a - 39b = 13(1 + 2a - 3b)$

(c) Common factor is $5x$ . $5xy \div 5x = y$ , and dividing $-5x$ by $5x$ leaves $-1$ :
$5xy - 5x = 5x(y - 1)$

(a) $9y(1 + 3z)$ (b) $13(1 + 2a - 3b)$ (c) $5x(y - 1)$

Question 6: Extracting Negative Common Factors

Factorise each expression by extracting a negative common factor:
(a) $-20u - 35v - 15$
(b) $-12x - 18y$
(c) $-4ab - 8ac$

ShowHide solution

(a) $\text{HCF}(20, 35, 15) = 5$ . Extract $-5$ :
$-20u - 35v - 15 = -5(4u + 7v + 3)$

(b) $\text{HCF}(12, 18) = 6$ . Extract $-6$ :
$-12x - 18y = -6(2x + 3y)$

(c) Both terms are negative and share $4a$ . Extract $-4a$ :
$-4ab - 8ac = -4a(b + 2c)$

(a) $-5(4u + 7v + 3)$ (b) $-6(2x + 3y)$ (c) $-4a(b + 2c)$

Question 7: Three Terms with a Common Letter

Factorise each expression completely:
(a) $6ab + 9ac - 12a$
(b) $13ab - 26ac + 13ad$
(c) $10pq - 15pr + 25p$

ShowHide solution

(a) $\text{HCF}(6, 9, 12) = 3$ . Every term has $a$ . Common factor is $3a$ :
$6ab + 9ac - 12a = 3a(2b + 3c - 4)$

(b) $\text{HCF}(13, 26, 13) = 13$ . Every term has $a$ . Common factor is $13a$ :
$13ab - 26ac + 13ad = 13a(b - 2c + d)$

(c) $\text{HCF}(10, 15, 25) = 5$ . Every term has $p$ . Common factor is $5p$ :
$10pq - 15pr + 25p = 5p(2q - 3r + 5)$

(a) $3a(2b + 3c - 4)$ (b) $13a(b - 2c + d)$ (c) $5p(2q - 3r + 5)$

Question 8: Factorise, Then Check by Expanding

Factorise each expression completely, then check your answer by expanding it:
(a) $-6p + 9q - 3$ , taking out a negative common factor
(b) $20xy - 30xz + 10x$

ShowHide solution

(a) $\text{HCF}(6, 9, 3) = 3$ . The first term is negative, so extract $-3$ . Dividing by $-3$ changes every sign:
$-6p \div (-3) = 2p, \quad 9q \div (-3) = -3q, \quad -3 \div (-3) = 1$
$-6p + 9q - 3 = -3(2p - 3q + 1)$
Check: $(-3) \times 2p + (-3) \times (-3q) + (-3) \times 1 = -6p + 9q - 3$ .

(b) $\text{HCF}(20, 30, 10) = 10$ , and every term has $x$ . Common factor is $10x$ ; $10x \div 10x = 1$ :
$20xy - 30xz + 10x = 10x(2y - 3z + 1)$
Check: $10x \times 2y - 10x \times 3z + 10x \times 1 = 20xy - 30xz + 10x$ .

(a) $-3(2p - 3q + 1)$ (b) $10x(2y - 3z + 1)$

Question 9: Rectangular Banner Dimensions in a Geometry Setting

The area of a rectangular banner is $(60x + 90)\text{ cm}^2$ , and its width is $30\text{ cm}$ .
(a) Factorise the area expression to find an expression for the length of the banner in terms of $x$ .
(b) Write down an expression for the perimeter of the banner in terms of $x$ .
(c) If $x = 50$ , calculate the length and the perimeter of the banner in centimetres.

ShowHide solution

(a) Area $=$ width $\times$ length. The HCF of $60$ and $90$ is $30$ , which is the width:
$60x + 90 = 30(2x + 3)$
Therefore, the length of the banner is $(2x + 3)\text{ cm}$ .

(b) Perimeter $= 2(\text{width} + \text{length})$ :
$\text{Perimeter} = 2[30 + (2x + 3)] = 2[2x + 33] = (4x + 66)\text{ cm}$

(c) Substitute $x = 50$ :
$\text{Length} = 2(50) + 3 = 100 + 3 = 103\text{ cm}$
$\text{Perimeter} = 4(50) + 66 = 200 + 66 = 266\text{ cm}$
Check perimeter: $2(30 + 103) = 2(133) = 266\text{ cm}$ .
Check area: $30 \times 103 = 3090\text{ cm}^2$ ; formula: $60(50) + 90 = 3000 + 90 = 3090\text{ cm}^2$ .

(a) $(2x + 3)\text{ cm}$ (b) $(4x + 66)\text{ cm}$ (c) Length $= 103\text{ cm}$ , Perimeter $= 266\text{ cm}$

Question 10: Square Picture Frame in a Setting

A picture framer makes a square wooden frame with a total perimeter of $(16w + 24)\text{ cm}$ .
(a) Factorise the perimeter expression completely by extracting its highest common factor.
(b) A square has four equal sides. Use your answer to (a) to find an expression for the length of one side in terms of $w$ .
(c) If $w = 3$ , calculate the side length and the perimeter in centimetres.

ShowHide solution

(a) The HCF of $16$ and $24$ is $8$ :
$16w + 24 = 8(2w + 3)$
The perimeter is $8(2w + 3)\text{ cm}$ .

(b) Divide the perimeter by $4$ . Since $8 \div 4 = 2$ , the factorised form makes this one step:
$\text{Side length} = 8(2w + 3) \div 4 = 2(2w + 3) = (4w + 6)\text{ cm}$

(c) Substitute $w = 3$ :
$\text{Side length} = 4(3) + 6 = 12 + 6 = 18\text{ cm}$
$\text{Perimeter} = 16(3) + 24 = 48 + 24 = 72\text{ cm}$
Check: $4 \times 18 = 72\text{ cm}$ .

(a) $8(2w + 3)\text{ cm}$ (b) $(4w + 6)\text{ cm}$ (c) Side length $= 18\text{ cm}$ , Perimeter $= 72\text{ cm}$

Question 11: Shelves of Shuttlecocks in a Setting

A sports shop has $18$ boxes of feather shuttlecocks with $p$ shuttlecocks in each box, and $30$ boxes of plastic shuttlecocks with $q$ shuttlecocks in each box. So it has $(18p + 30q)$ shuttlecocks in all.
(a) Factorise $18p + 30q$ completely.
(b) The shopkeeper puts all the boxes on some shelves, so that every shelf holds the same number of feather boxes as every other shelf, and the same number of plastic boxes as every other shelf. What is the greatest number of shelves that can be used, and how many shuttlecocks, in terms of $p$ and $q$ , are on each shelf?
(c) If $p = 12$ and $q = 6$ , and each shuttlecock sells for $2, find the number of shuttlecocks on each shelf and the total amount received when all of them are sold.

ShowHide solution

(a) The HCF of $18$ and $30$ is $6$ :
$18p + 30q = 6(3p + 5q)$

(b) The number of shelves must divide both $18$ (the feather boxes) and $30$ (the plastic boxes) exactly. The greatest such number is $\text{HCF}(18, 30) = 6$ .
Each shelf then has $18 \div 6 = 3$ boxes of feather shuttlecocks and $30 \div 6 = 5$ boxes of plastic shuttlecocks, which is $(3p + 5q)$ shuttlecocks: the bracket in (a).

(c) Substitute $p = 12$ and $q = 6$ :
$\text{On each shelf} = 3(12) + 5(6) = 36 + 30 = 66$
$\text{Total} = 18(12) + 30(6) = 216 + 180 = 396$
Check: $6 \times 66 = 396$ .
At $2 each, the total amount received is $396 \times 2 = 792$ , so $792.

(a) $6(3p + 5q)$ (b) $6$ shelves, each with $(3p + 5q)$ shuttlecocks (c) $66$ shuttlecocks on each shelf; $792

Question 12: Community Hall Flooring in a Setting

A community hall floor consists of two rectangular sections:

Section A has area $(15k + 10)\text{ m}^2$ .
Section B has area $(20k + 15)\text{ m}^2$ .
(a) Write down and simplify an expression for the total area of the floor in terms of $k$ .
(b) Factorise the total area expression completely by extracting its highest common factor.
(c) If $k = 10$ , calculate the area of Section A, Section B, and the total floor area in square metres.

ShowHide solution

(a) Sum the areas of both sections:
$\begin{aligned} \text{Total area} &= (15k + 10) + (20k + 15) \\ &= (35k + 25)\text{ m}^2 \end{aligned}$

(b) The HCF of $35$ and $25$ is $5$ :
$35k + 25 = 5(7k + 5)$
The total area is $5(7k + 5)\text{ m}^2$ . Inside the bracket, $7$ and $5$ have no common factor other than $1$ , so the factorisation is complete.

(c) Substitute $k = 10$ :
$\text{Section A} = 15(10) + 10 = 150 + 10 = 160\text{ m}^2$
$\text{Section B} = 20(10) + 15 = 200 + 15 = 215\text{ m}^2$
$\text{Total area} = 160 + 215 = 375\text{ m}^2$
Check with the factorised form: $5(7(10) + 5) = 5(75) = 375\text{ m}^2$ .

(a) $(35k + 25)\text{ m}^2$ (b) $5(7k + 5)\text{ m}^2$ (c) Section A: $160\text{ m}^2$ , Section B: $215\text{ m}^2$ , Total: $375\text{ m}^2$

Key Points to Remember

Factorisation is the Reverse of Expansion: Writing an algebraic expression as a product of factors: $6x + 9 = 3(2x + 3)$ .
Find the Complete HCF: Determine the highest common factor of all the numbers in the terms (the coefficients and any constant term) and every letter common to all the terms.
Complete Factorisation: Ensure the terms remaining inside the bracket share no common factors other than $1$ .
The Vanishing Term Trap: When an entire term is taken out, it leaves behind $1$ , not $0$ ( $11a \div 11a = 1$ ).
Extracting Negative Factors: When the first term is negative, it is often neater to take out a negative factor: $-4x - 6 = -2(2x + 3)$ .
The Unit Rule: When an algebraic expression has more than one term and carries a unit, always enclose the expression in brackets, such as $(2x + 3)\text{ cm}$ or $(3p + 4)$ dollars.