How To Factorise Algebraic Expressions: A Step-By-Step Tutorial For Singapore Secondary Students

Factorisation is one of those topics that keeps coming back in Secondary Maths.

Sec 1: simple common factors.
Sec 2: quadratics and special formulas.
Sec 3–4: harder O-Level questions mixed with equations, algebraic fractions, and word problems.

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If you’re in a Singapore secondary school, you must be solid at factorisation. It appears in:

• Sec 1–2 E-Math topics
• Sec 3–4 E-Math and A-Math
• N Level and O Level papers

In this tutorial, I’ll walk you through how to factorise algebraic expressions step by step, the way your MOE teacher expects, plus exam strategies and practice.

Throughout, I’ll also show you how to use Tutorly.sg, a 24/7 AI tutor built specifically for the Singapore MOE syllabus, to drill factorisation efficiently. Tutorly.sg has already been used by thousands of students in Singapore and has even been mentioned on CNA (Channel NewsAsia), so you’re in good hands.

You can try the AI tutor here:
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Step-by-step tutorial

Let’s go from the basics up to the harder types you’ll see in O Levels.

1. Factorising by common factor

This is the simplest and appears in Sec 1 and early Sec 2.

Idea: Find the greatest common factor (GCF) of all terms and factor it out.

Example 1
Factorise: $6 x^2 y - 9xy^2$

1. Look at numbers: $\gcd(6, 9) = 3$
2. Look at $x$: $\min(2, 1) = 1$ so common factor has $x^1$
3. Look at $y$: $\min(1, 2) = 1$ so common factor has $y^1$

So common factor is $3xy$.

4. Factor out $3xy$:

$6 x^2 y - 9xy^2 = 3xy(2 x - 3 y)$

Example 2
Factorise: $5 a^3 b^2 + 10 a^2 b$

1. GCF of numbers: $\gcd(5,10) = 5$
2. $a$: $\min(3,2) = 2$
3. $b$: $\min(2,1) = 1$

Common factor: $5 a^2 b$

$5 a^3 b^2 + 10 a^2 b = 5 a^2 b(a b + 2)$

What to remember:

• Always check numbers and letters.
• Take the smallest power of each letter that appears in all terms.

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2. Factorising by grouping

This is usually in Sec 2 or early Sec 3. It appears often in O-Level Paper 1.

Idea: Group the terms into pairs, factor each pair, then factor again.

Example 3
Factorise: $ax + ay + bx + by$

1. Group: $(ax + ay) + (bx + by)$

2. Factor each group:

   • $ax + ay = a(x + y)$
   • $bx + by = b(x + y)$

3. Now factor $(x + y)$:

$ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)$

Example 4
Factorise: $3 p^2 - 6 p + 2 p - 4$

1. Group: $(3 p^2 - 6 p) + (2 p - 4)$

2. Factor each group:

   • $3 p^2 - 6 p = 3 p(p - 2)$
   • $2 p - 4 = 2(p - 2)$

3. Factor $(p - 2)$:

$3 p^2 - 6 p + 2 p - 4 = 3 p(p - 2) + 2(p - 2) = (3 p + 2)(p - 2)$

Tips:

• Rearrange terms if needed so that grouping works.
• Always look for a common bracket after the first round of factorisation.

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3. Special products: $(a + b)^2$, $(a - b)^2$, $a^2 - b^2$

These are standard formulas in the MOE syllabus and very popular in exams.

3.1 Perfect square trinomials

You must recognise:

• $(a + b)^2 = a^2 + 2ab + b^2$
• $(a - b)^2 = a^2 - 2ab + b^2$

Example 5
Factorise: $x^2 + 6 x + 9$

1. Check first term: $x^2 = x^2$
2. Check last term: $9 = 3^2$
3. Middle term: $6 x = 2 \cdot x \cdot 3$

So it matches $a^2 + 2ab + b^2$ with $a = x$, $b = 3$:

$x^2 + 6 x + 9 = (x + 3)^2$

Example 6
Factorise: $4 y^2 - 12 y + 9$

1. $4 y^2 = (2 y)^2$
2. $9 = 3^2$
3. Middle term: $-12 y = -2 \cdot (2 y) \cdot 3$

So it matches $(a - b)^2$ with $a = 2 y$, $b = 3$:

$4 y^2 - 12 y + 9 = (2 y - 3)^2$

3.2 Difference of two squares

Formula:
$a^2 - b^2 = (a + b)(a - b)$

Example 7
Factorise: $x^2 - 25$

$x^2 = x^2$, $25 = 5^2$

$x^2 - 25 = (x + 5)(x - 5)$

Example 8
Factorise: $9 p^2 - 16 q^2$

$9 p^2 = (3 p)^2$, $16 q^2 = (4 q)^2$

$9 p^2 - 16 q^2 = (3 p + 4 q)(3 p - 4 q)$

Important:
$a^2 + b^2$ cannot be factorised over real numbers at your level. Don’t force it.

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4. Quadratic trinomials: $ax^2 + bx + c$

This is core Sec 3–4 content and very common in N/O Level.

There are two main cases:

1. $x^2 + bx + c$ (coefficient of $x^2$ is 1)
2. $ax^2 + bx + c$ (coefficient of $x^2$ is not 1)

4.1 Case 1: $x^2 + bx + c$

Method: Find two numbers that multiply to $c$ and add to $b$.

Example 9
Factorise: $x^2 + 7 x + 12$

1. Multiply to $12$, add to $7$:

   • $3 \cdot 4 = 12$ and $3 + 4 = 7$

2. So:

$x^2 + 7 x + 12 = (x + 3)(x + 4)$

Example 10
Factorise: $x^2 - 5 x + 6$

1. Multiply to $6$, add to $-5$:

   • $-2 \cdot -3 = 6$, and $-2 + -3 = -5$

2. So:

$x^2 - 5 x + 6 = (x - 2)(x - 3)$

4.2 Case 2: $ax^2 + bx + c$ with $a \neq 1$

There are different methods. The most exam-friendly for many students is splitting the middle term.

Example 11
Factorise: $2 x^2 + 7 x + 3$

1. Multiply $a \cdot c = 2 \cdot 3 = 6$

2. Find two numbers that multiply to $6$ and add to $7$:

   • $1$ and $6$

3. Split $7 x$ into $1 x + 6 x$:

$2 x^2 + 7 x + 3 = 2 x^2 + x + 6 x + 3$

4. Group:

$(2 x^2 + x) + (6 x + 3)$

5. Factor each group:

• $2 x^2 + x = x(2 x + 1)$
• $6 x + 3 = 3(2 x + 1)$

6. Factor $(2 x + 1)$:

$2 x^2 + 7 x + 3 = x(2 x + 1) + 3(2 x + 1) = (x + 3)(2 x + 1)$

Example 12
Factorise: $3 x^2 - 8 x - 3$

1. $a \cdot c = 3 \cdot (-3) = -9$

2. Need two numbers that multiply to $-9$ and add to $-8$:

   • $1$ and $-9$ (since $1 - 9 = -8$)

3. Split $-8 x$:

$3 x^2 + x - 9 x - 3$

4. Group:

$(3 x^2 + x) + (-9 x - 3)$

5. Factor each group:

• $3 x^2 + x = x(3 x + 1)$
• $-9 x - 3 = -3(3 x + 1)$

6. Factor $(3 x + 1)$:

$3 x^2 - 8 x - 3 = x(3 x + 1) - 3(3 x + 1) = (x - 3)(3 x + 1)$

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5. Factorising expressions with more than one variable

These appear in mid–upper secondary and O-Level questions.

Example 13
Factorise: $x^2 y + 3xy^2$

1. Common factor: $xy$

$x^2 y + 3xy^2 = xy(x + 3 y)$

Example 14
Factorise: $2 a^2 b - 8ab^2 + 6ab$

1. Common factor: $2ab$

$2 a^2 b - 8ab^2 + 6ab = 2ab(a - 4 b + 3)$

2. Check if inside bracket can be factorised further: $a - 4 b + 3$
   • This is linear in $a$ and $b$; no more factorisation.

Sometimes you’ll get quadratics in one variable but with another variable as a constant.

Example 15
Factorise: $x^2 + (k - 3)x - 4 k$

Treat $k$ as a constant. We want two numbers that:

• Multiply to $-4 k$
• Add to $(k - 3)$

Try $4$ and $-k$:

• $4 \cdot (-k) = -4 k$
• $4 + (-k) = 4 - k$ (not $k - 3$)

Try $(k + 1)$ and $(-4)$:

• $(k + 1)(-4) = -4 k - 4$ (not $-4 k$)

This one is trickier; in exams, such questions are usually crafted to be factorisable nicely, or they ask you to show something. Don’t panic; just systematically test pairs or use Tutorly.sg to check if your factorisation is correct.

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6. Factorising algebraic fractions (briefly)

At O Level, factorisation is often used to simplify algebraic fractions.

Example 16
Simplify: $\dfrac{x^2 - 9}{x^2 - 6 x + 9}$

1. Factor numerator: $x^2 - 9 = (x + 3)(x - 3)$
2. Factor denominator: $x^2 - 6 x + 9 = (x - 3)^2$

So:

$\dfrac{x^2 - 9}{x^2 - 6 x + 9} = \dfrac{(x + 3)(x - 3)}{(x - 3)(x - 3)} = \dfrac{x + 3}{x - 3}, \quad x \neq 3$

The key is still factorisation; the fraction part is just cancellation.

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Exam strategy guide

Knowing how to factorise isn’t enough. In exams $especially O Level E-Math Paper 1$, you must be fast and accurate.

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Here are strategies specific to Singapore secondary students.

1. When to factorise in exam questions

Look out for phrases like:

• “Factorise …” (obvious)
• “Hence solve the equation …”
• “Simplify the expression …”
• “Solve the quadratic equation …”

In many solve questions, you’re expected to:

1. Factorise the expression
2. Use each factor to find $x$

Example 17
Solve: $x^2 - 5 x + 6 = 0$

1. Factorise: $(x - 2)(x - 3) = 0$
2. So $x - 2 = 0$ or $x - 3 = 0$
3. $x = 2$ or $x = 3$

2. Time management for factorisation questions

In O-Level E-Math Paper 1:

• Short factorisation questions $2–3 marks$ should take under 3 minutes.
• If you are stuck for more than 3 minutes on a single factorisation, move on and come back later.

Use practice to build pattern recognition:

• See $x^2 + 2xy + y^2$ → think $(x + y)^2$
• See $a^2 - 16$ → think $(a + 4)(a - 4)$
• See $3 x^2 - 8 x - 3$ → think “split middle term”

Using Tutorly.sg, you can quickly try many similar questions and see worked solutions, which helps your brain get used to common patterns.

Try it here: https://tutorly.sg/ai-tutor-singapore

3. Showing working clearly (MOE marking style)

Markers want to see:

1. A clear factorised form
2. Logical steps if the question is more than 1 mark

Example (2-mark question):
Factorise completely: $3 x^2 - 3$

A good working:

$3 x^2 - 3 = 3(x^2 - 1) = 3(x + 1)(x - 1)$

Don’t jump straight to $3(x + 1)(x - 1)$ if you’re unsure; show the intermediate step.

4. Using “hence” in combined questions

Common format:

1. (a) Factorise some expression
2. (b) Hence solve / simplify something else

The “hence” means you should reuse your factorised form from part (a). This saves time and is usually easier.

If you miss the “hence” and start from scratch, you might waste precious minutes.

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Worksheet practice

Use this section like a mini worksheet. Try each question first, then compare with the worked solution.

To get more questions instantly, you can open Tutorly.sg in another tab and ask it for “Sec 3 E-Math factorisation practice” or “O Level factorisation worksheet”. It will generate questions aligned to the MOE syllabus and show step-by-step solutions after you submit your final answer.

👉 https://tutorly.sg/app

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A. Basic to intermediate practice

Q 1

Factorise: $8 x^3 y^2 - 4 x^2 y$

Solution:

Common factor: $4 x^2 y$

$8 x^3 y^2 - 4 x^2 y = 4 x^2 y(2xy - 1)$

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Q 2

Factorise: $5 a^2 - 20 a$

Solution:

Common factor: $5 a$

$5 a^2 - 20 a = 5 a(a - 4)$

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Q 3

Factorise: $x^2 - 10 x + 25$

Solution:

Recognise perfect square:

• $x^2 = x^2$
• $25 = 5^2$
• Middle term $-10 x = -2 \cdot x \cdot 5$

So:

$x^2 - 10 x + 25 = (x - 5)^2$

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Q 4

Factorise: $9 y^2 - 1$

Solution:

Difference of squares:

• $9 y^2 = (3 y)^2$
• $1 = 1^2$

$9 y^2 - 1 = (3 y + 1)(3 y - 1)$

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Q 5

Factorise: $x^2 + 2xy + y^2$

Solution:

Recognise: $(x + y)^2$

$x^2 + 2xy + y^2 = (x + y)^2$

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B. Quadratic factorisation practice

Q 6

Factorise: $x^2 + 11 x + 24$

Solution:

Need numbers that multiply to $24$ and add to $11$:

• $3$ and $8$

So:

$x^2 + 11 x + 24 = (x + 3)(x + 8)$

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Q 7

Factorise: $x^2 - x - 12$

Solution:

Multiply to $-12$, add to $-1$:

• $3$ and $-4$ (since $3 \cdot -4 = -12$ and $3 + -4 = -1$)

So:

$x^2 - x - 12 = (x + 3)(x - 4)$

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Q 8

Factorise: $2 x^2 + 5 x - 3$

Solution:

1. $a \cdot c = 2 \cdot (-3) = -6$

2. Need two numbers multiply to $-6$ and add to $5$:

   • $6$ and $-1$

3. Split middle term:

$2 x^2 + 6 x - x - 3$

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4. Group:

$(2 x^2 + 6 x) + (-x - 3)$

5. Factor each group:

• $2 x^2 + 6 x = 2 x(x + 3)$
• $-x - 3 = -1(x + 3)$

6. Factor $(x + 3)$:

$2 x^2 + 5 x - 3 = 2 x(x + 3) - 1(x + 3) = (2 x - 1)(x + 3)$

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Q 9

Factorise: $3 x^2 + 10 x + 3$

Solution:

1. $a \cdot c = 3 \cdot 3 = 9$

2. Need two numbers multiply to $9$ and add to $10$:

   • $1$ and $9$

3. Split:

$3 x^2 + x + 9 x + 3$

4. Group:

$(3 x^2 + x) + (9 x + 3)$

5. Factor:

• $3 x^2 + x = x(3 x + 1)$
• $9 x + 3 = 3(3 x + 1)$

6. Factor $(3 x + 1)$:

$3 x^2 + 10 x + 3 = (x + 3)(3 x + 1)$

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C. Harder exam-style variants

These are closer to what you might see in Sec 3–4 tests or O-Level papers.

Q 10

Factorise completely: $6 x^2 - 24$

Solution:

1. Factor out common factor:

$6 x^2 - 24 = 6(x^2 - 4)$

2. Recognise difference of squares:

$x^2 - 4 = (x + 2)(x - 2)$

3. Final answer:

$6 x^2 - 24 = 6(x + 2)(x - 2)$

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Q 11

Factorise completely: $x^3 - x$

Solution:

1. Common factor: $x$

$x^3 - x = x(x^2 - 1)$

2. Difference of squares:

$x^2 - 1 = (x + 1)(x - 1)$

3. Final:

$x^3 - x = x(x + 1)(x - 1)$

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Q 12

Factorise completely: $4 x^2 - 9 y^2 + 12xy$

This one is not in standard order. Rearranging might help:

$4 x^2 + 12xy - 9 y^2$

Try to see if it’s a perfect square trinomial:

• $4 x^2 = (2 x)^2$
• $9 y^2 = (3 y)^2$
• Middle term for $(2 x + 3 y)^2$ would be $2 \cdot 2 x \cdot 3 y = 12xy$

So:

$4 x^2 + 12xy - 9 y^2 = (2 x + 3 y)^2 - (3 y)^2 - 9 y^2$

But that’s messy. Instead, think:

$4 x^2 + 12xy + 9 y^2 - 18 y^2$

$= (2 x + 3 y)^2 - (3\sqrt{2}y)^2$

That’s not nice either. Let’s try another approach: treat as quadratic in $x$.

Write as:

$4 x^2 + 12xy - 9 y^2$

Think of $x$ as the variable, $y$ as constant:

1. Coefficients: $a =

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