Factorization is the bedrock of DSE algebra. This article covers the three essential techniques every candidate must master: taking out the common factor, applying algebraic identities (difference of squares, perfect squares, sum/difference of cubes), and grouping. You will learn how to recognise which method to use, avoid common traps, and apply these skills to DSE-style Section A(1) and A(2) questions. Complete with worked examples, exam tips, and 10 practice questions.
Learning Objectives
- Identify and extract the highest common factor (HCF) from algebraic terms.
- Apply the three key identities: \(a^2-b^2=(a+b)(a-b)\), \(a^2\pm 2ab+b^2=(a\pm b)^2\), and \(a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)\).
- Use the grouping method to factorise four-term expressions.
- Combine techniques to factorise more complex polynomials.
- Apply these skills to DSE Paper 1 Section A and Paper 2 MC questions with speed and accuracy.
Introduction
Factorization is the process of rewriting an algebraic expression as a product of simpler factors.
In the DSE Mathematics Compulsory Part, factorization appears not only in pure algebra questions but
also in solving equations, simplifying fractions, and even in coordinate geometry. The HKEAA syllabus
requires candidates to be fluent in three core techniques:
- Common factor – the first step in almost every factorization.
- Formulas (identities) – using standard algebraic patterns.
- Grouping – a strategic method for four or more terms.
In this article, we break down each technique with clear steps, worked examples, and DSE-style practice.
By the end, you will be able to tackle any factorization question in Section A(1) and A(2) confidently.
Technique 1: Taking Out the Common Factor
The most fundamental step: look for a factor that divides every term of the expression.
This factor can be a number, a variable, or a combination of both.
DSE Exam Tip
Always check for a common factor first. Many DSE students lose easy marks by skipping this step.
Even if you plan to use another method, extracting the common factor first simplifies the expression and
reduces errors.
Steps
- Identify the greatest common factor (GCF) of the coefficients.
- Identify the lowest power of each variable that appears in all terms.
- Write the common factor outside the brackets, and divide each term by it.
Worked Example 1
Example 1
Factorise \(12x^3y^2 - 18x^2y^3 + 6x^2y^2\).
Solution:
- Coefficients: GCF of 12, 18, 6 is 6.
- Variables: \(x^2\) is the lowest power of \(x\) (since \(x^2\) appears in all terms), and \(y^2\) is the lowest power of \(y\).
- Common factor: \(6x^2y^2\).
\[
12x^3y^2 - 18x^2y^3 + 6x^2y^2
= 6x^2y^2(2x - 3y + 1)
\]
Answer: \(6x^2y^2(2x - 3y + 1)\)
Technique 2: Using Algebraic Identities (Formulas)
There are three key identities you must memorize for the DSE. They allow you to factorise
special binomials and trinomials instantly.
Memory Aid
Think of the identities as "patterns to spot":
difference of two squares, perfect square trinomials, and sum/difference of cubes.
Practice recognising them by looking at the structure of the expression.
Identity 1: Difference of Two Squares
\[
a^2 - b^2 = (a + b)(a - b)
\]
Identity 2: Perfect Square Trinomials
\[
a^2 + 2ab + b^2 = (a + b)^2,\qquad a^2 - 2ab + b^2 = (a - b)^2
\]
Identity 3: Sum / Difference of Cubes
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2),\qquad
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]
Common Trap
Do not confuse \(a^2 + b^2\) with \((a+b)^2\). There is no simple factorisation
for \(a^2 + b^2\) over the real numbers. Also, remember that for difference of cubes,
the second factor is \(a^2 + ab + b^2\) (not \(a^2 - ab + b^2\)).
Worked Example 2
Example 2
Factorise \(4x^2 - 12x + 9\).
Solution:
This is a perfect square trinomial because \(4x^2 = (2x)^2\), \(9 = 3^2\), and the middle term
\(-12x = -2 \cdot (2x) \cdot 3\).
\[
4x^2 - 12x + 9 = (2x - 3)^2
\]
Answer: \((2x - 3)^2\)
Worked Example 3
Example 3
Factorise \(27x^3 - 8\).
Solution:
This is a difference of cubes: \((3x)^3 - 2^3\).
Using \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) with \(a=3x,\ b=2\):
\[
27x^3 - 8 = (3x - 2)\bigl((3x)^2 + (3x)(2) + 2^2\bigr)
= (3x - 2)(9x^2 + 6x + 4)
\]
Answer: \((3x - 2)(9x^2 + 6x + 4)\)
Technique 3: Grouping Method
When an expression has four (or more) terms and no common factor applies to all terms,
grouping can be the key. The idea is to group terms in pairs so that each pair has a common factor,
and then factor out the common binomial.
Steps
- Group the terms into two pairs (usually first two and last two).
- Factor out the common factor from each pair.
- If the two resulting groups share a common binomial factor, factor it out.
Worked Example 4
Example 4
Factorise \(2x^3 - 3x^2 + 4x - 6\).
Solution:
Group the terms:
\[
(2x^3 - 3x^2) + (4x - 6)
\]
Factor each pair:
\[
x^2(2x - 3) + 2(2x - 3)
\]
Now factor out the common binomial \((2x - 3)\):
\[
(2x - 3)(x^2 + 2)
\]
Answer: \((2x - 3)(x^2 + 2)\)
DSE Exam Tip
If the first grouping doesn't work, try re-grouping. Sometimes swapping the middle terms
or grouping in a different way reveals a common factor. Practice flexibility!
Combining Techniques
In the DSE, you will often need to apply more than one technique. The general strategy is:
- Common factor first – always extract the GCF before anything else.
- Look for identities – difference of squares, perfect squares, or cubes.
- Try grouping if the expression has four or more terms.
- Repeat until all factors are irreducible over the integers.
Worked Example 5 (Combined)
Example 5
Factorise \(2x^4 - 2x^2 - 4x^3 + 4x\) completely.
Solution:
Step 1 – Common factor: \(2x\) is common to all terms.
\[
2x(x^3 - 2x^2 - x + 2)
\]
Step 2 – Group the cubic inside:
\[
x^3 - 2x^2 - x + 2 = (x^3 - 2x^2) + (-x + 2)
= x^2(x - 2) - 1(x - 2)
= (x - 2)(x^2 - 1)
\]
Step 3 – Apply difference of squares to \(x^2 - 1 = (x - 1)(x + 1)\).
Therefore:
\[
2x^4 - 2x^2 - 4x^3 + 4x
= 2x(x - 2)(x - 1)(x + 1)
\]
Answer: \(2x(x - 2)(x - 1)(x + 1)\)
DSE Exam Techniques – Paper 1 & Paper 2
Paper 1 Section A(1) & A(2)
- Show all steps – you get method marks even if your final answer is wrong.
- Always state the common factor you are extracting.
- For identities, explicitly mention the formula you are using (e.g. "difference of squares").
- Check your answer by expanding – a quick expansion verifies your factorization.
Paper 2 (Multiple Choice)
- Spot the pattern quickly: look at the first and last terms to guess the identity.
- Use the answer options to guide you – if the options are factors, test one factor by substitution.
- For grouping questions, try pairing terms in the order they appear, but if it doesn't work, re-group mentally.
DSE-Style Practice Questions
Try these questions. Full solutions are provided below.
Practice Set 1 – Common Factor
- \(6x^2y - 9xy^2 + 3xy\)
- \(4a^3b^2 - 8a^2b^3 + 12a^2b^2\)
- \(15p^2q^3 - 5p q^2 + 25p^3 q^2\)
Practice Set 2 – Identities
- \(9x^2 - 16\)
- \(25x^2 - 30x + 9\)
- \(8x^3 + 27\)
- \(64x^3 - 1\)
Practice Set 3 – Grouping
- \(xy + 3x + 2y + 6\)
- \(2x^3 - 4x^2 + 3x - 6\)
- \(a^2 - b^2 + a - b\) Harder
Practice Set 4 – Combined Techniques
- \(3x^3 - 12x\)
- \(2x^4 - 2\)
- \(x^4 - 5x^2 + 4\) Challenging
Solutions to Practice Sets
Set 1 Solutions
- \(3xy(2x - 3y + 1)\)
- \(4a^2b^2(a - 2b + 3)\)
- \(5p q^2(3p q - 1 + 5p^2)\)
Set 2 Solutions
- \((3x - 4)(3x + 4)\)
- \((5x - 3)^2\)
- \((2x + 3)(4x^2 - 6x + 9)\)
- \((4x - 1)(16x^2 + 4x + 1)\)
Set 3 Solutions
- \((x + 2)(y + 3)\)
- \((x - 2)(2x^2 + 3)\)
-
\(a^2 - b^2 + a - b = (a - b)(a + b) + (a - b) = (a - b)(a + b + 1)\)
Set 4 Solutions
- \(3x(x^2 - 4) = 3x(x - 2)(x + 2)\)
- \(2(x^4 - 1) = 2(x^2 - 1)(x^2 + 1) = 2(x - 1)(x + 1)(x^2 + 1)\)
-
\(x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4) = (x - 1)(x + 1)(x - 2)(x + 2)\)
9. Exercise
Click the following link to have
An Exercise on Three Core Factorization Techniques: Common Factor + Formulas + Grouping
Summary Checklist for Revision
-
Common factor: I always extract the GCF before applying other methods.
-
Difference of squares: I can spot \(a^2 - b^2\) and factorise it immediately.
-
Perfect squares: I recognise \(a^2 \pm 2ab + b^2\) and write it as \((a \pm b)^2\).
-
Cubes: I memorised the sum and difference of cubes formulas and use them correctly.
-
Grouping: I can group four-term expressions and factor out common binomials.
-
Combined: I use common factor first, then identities or grouping, and factor completely.
-
Verification: I always expand my answer to check (mentally or on paper).