Introduction to Factorisation

Common factors method and factorisation by grouping.

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What is Factorisation?

Factorisation is the process of expressing a number or an algebraic expression as a product of its factors. Just as 12 can be written as 3 × 4 or 2 × 6, algebraic expressions like 6x² + 9xy can be rewritten as a product of simpler parts. This chapter — Chapter 12 of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) — teaches you the key techniques to do exactly that.

The introduction begins with a familiar idea — the factors of the number 36 — and builds up to factorising algebraic expressions using the Common Factor Method and the Grouping Method. Mastering this introduction sets the stage for all the exercises in Chapter 12.

Factors of Numbers Prime Factorisation Irreducible Factors HCF Method Grouping Method

Core Idea of Chapter 12

Expression = Factor₁ × Factor₂ × ...

Breaking an expression into a product of its simplest (irreducible) parts — just like splitting 36 into 2 × 2 × 3 × 3

Starting Point — How Many Ways Can 36 Be Written as a Product?

The textbook opens Chapter 12 with a simple but powerful question about the number 36. How many different ways can 36 be written as a product of two numbers? The answers reveal the concept of factors clearly.

Product Form	Factor 1	Factor 2	Valid?
36 = 1 × 36	1	36	✅ Yes
36 = 2 × 18	2	18	✅ Yes
36 = 3 × 12	3	12	✅ Yes
36 = 4 × 9	4	9	✅ Yes
36 = 6 × 6	6	6	✅ Yes

So the factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36 — all the whole numbers that divide 36 exactly without leaving a remainder.

💡 Definition: A factor of a number is any number that divides it completely (i.e., with remainder 0). Every number has at least two factors: 1 and itself.

Prime Factorisation — Breaking 36 Down to Its Prime Factors

No matter which starting product you choose, if you keep splitting the factors until only prime numbers remain, you always end up with the same set of primes. This is the Fundamental Theorem of Arithmetic: every number has a unique prime factorisation.

The textbook shows four students — Alekhya, Asifa, Fariya, and Sri Laxmi — each starting from a different product of 36, but all reaching the same prime factorisation:

Alekhya

36 = 2 × 18
= 2 × 2 × 9
= 2 × 2 × 3 × 3
= 2² × 3²

Asifa

36 = 3 × 12
= 3 × 2 × 6
= 3 × 2 × 2 × 3
= 2² × 3²

Fariya

36 = 4 × 9
= 2 × 2 × 3 × 3

= 2² × 3²

Sri Laxmi

36 = 6 × 6
= 2 × 3 × 2 × 3

= 2² × 3²

✅ Key Conclusion: No matter how you start, 36 always breaks down to 2 × 2 × 3 × 3. Any number can be expressed uniquely as a product of prime factors — this is called Prime Factorisation.

36 = 2² × 3² ← the unique prime factorisation of 36

Factors of Algebraic Expressions — Irreducible Factors

The same idea of breaking into factors extends to algebraic expressions. Just as 36 = 2 × 2 × 3 × 3, an algebraic term like 3ab can be written as 3 × a × b. Each of these parts — 3, a, and b — cannot be broken down any further. Such factors are called irreducible factors.

Irreducible Factors of Algebraic Terms

📌 What makes a factor "irreducible"? A factor is irreducible when it cannot be split into a product of smaller factors. Numbers like 2, 3, 5, 7 are irreducible. Variables like a, x are irreducible. But 6 is NOT irreducible — it splits into 2 × 3. A bracket like (y + z) is treated as a single irreducible unit.

Example 1

Find the irreducible factors of 3ab

3ab = 3 × a × b Irreducible factors: 3, a, b None of these can be factored further — 3 is prime, a and b are variables

Example 2

Find the irreducible factors of 14x(y + z)

14x(y+z) = 2 × 7 × x × (y+z) Irreducible factors: 2, 7, x, (y+z) 14 = 2 × 7 (not prime, so split it) | (y+z) is kept as one unit

Why is Factorisation Needed? — Simplifying Algebraic Expressions

You might wonder: why go through the effort of factorising? The answer is that factorisation is the key tool for simplifying algebraic fractions — expressions where a polynomial is divided by another polynomial. Without factorisation, such divisions are impossible to simplify.

Real-Life Need for Factorisation

(x² + 7x + 12) ÷ (x + 3) = ?

You cannot simplify this until you factorise the numerator first!

Here is the step-by-step process the textbook demonstrates:

Worked Example — Division using Factorisation

Simplify: (x² + 7x + 12) ÷ (x + 3)

Step 1 — Write as a fraction

(x² + 7x + 12) ÷ (x + 3) = (x² + 7x + 12) / (x + 3)

Step 2 — Factorise the numerator

Split the middle term 7x as 3x + 4x (since 3 × 4 = 12 and 3 + 4 = 7):

x² + 7x + 12 = x² + 3x + 4x + 12 ← split 7x = x(x + 3) + 4(x + 3) ← group and take common factor = (x + 3)(x + 4) ← (x+3) is common

Step 3 — Cancel the common factor

(x² + 7x + 12) / (x + 3) = (x + 3)(x + 4) / (x + 3) ← (x+3) cancels = x + 4 ✓

✅ Without factorising the numerator, it is impossible to simplify this fraction. Factorisation is the essential first step.

Method 1 — Factorisation by Taking Out the Common Factor (HCF Method)

The most fundamental method of factorisation is to identify the Highest Common Factor (HCF) of all the terms in the expression, then "take it out" — writing the expression as HCF × (remaining bracket). This works whenever two or more terms share a common factor.

The HCF Method — How It Works

ax + ay = a(x + y)

Find the HCF of all terms → divide each term by HCF → put HCF outside, remainders inside the bracket

Full Worked Example from the Textbook: Factorise 6x² + 9xy

Common Factor Method

Factorise: 6x² + 9xy

Step 1 — Write prime factors of each term

Term 1: 6x²

6x² = 2 × 3 × x × x

Term 2: 9xy

9xy = 3 × 3 × x × y

Step 2 — Identify the common factors (HCF)

The factors appearing in both terms are 3 and x.

HCF of 6x², 9xy = 3 × x = 3x

Step 3 — Rewrite each term with the HCF factored out

6x² = (3x) × 2x ← 6x² ÷ 3x = 2x 9xy = (3x) × 3y ← 9xy ÷ 3x = 3y

Step 4 — Write the factorised form

6x² + 9xy = (3x)(2x) + (3x)(3y) = (3x)(2x + 3y) ← take 3x common = 3x(2x + 3y) ✓

✅ Verification: Expand back: 3x × 2x + 3x × 3y = 6x² + 9xy ✓ Always check your answer by expanding!

💡 Quick Rule: To factorise by common factors:
(1) Find HCF of all terms → (2) Divide each term by HCF → (3) Write: HCF × (quotients in bracket)

Method 2 — Factorisation by Grouping the Terms

Sometimes an expression has four or more terms and no single common factor for all of them. In such cases, we use the Grouping Method: pair the terms into groups so that each group has its own common factor. After factorising each group, a common binomial factor emerges that can be taken out.

Grouping Method

Factorise: x² + xy + x + y

Step 1 — Group the four terms into two pairs

x² + xy + x + y = (x² + xy) + (x + y) ← group first two and last two

Step 2 — Factorise each group separately

Group 1: (x² + xy)

= x · x + x · y
= x(x + y)

Group 2: (x + y)

= 1(x + y)

Step 3 — Identify the common binomial factor

= x(x + y) + 1(x + y) ← both groups have (x+y) = (x + y) · (x + 1) ← take (x+y) common = (x + y)(x + 1) ✓

✅ Verification: (x+y)(x+1) = x²+x+xy+y = x²+xy+x+y ✓

📌 Why does grouping work? The magic step is Step 3: after factorising each group, both results contain the same bracket — (x+y) in this case. This shared bracket is the common factor that can be pulled out, giving the final two-bracket answer.

Comparing the Two Methods Side by Side

Feature	Common Factor Method	Grouping Method
When to use	All terms share a common factor (number, variable, or both)	Four+ terms; no single common factor for all terms
Key step	Find HCF and take it outside	Group terms in pairs; factor each group; find common binomial
Example	6x² + 9xy → 3x(2x + 3y)	x² + xy + x + y → (x+y)(x+1)
Result form	Monomial × Polynomial	Binomial × Binomial
Check by	Expanding: distribute the monomial back	Expanding: FOIL or distributive law

How to Approach Any Factorisation Problem — A Simple Checklist

Count the terms. Two or three terms? Try the Common Factor (HCF) method first. Four or more terms? Grouping is usually the answer.

Find the HCF of all terms by writing out prime factors of coefficients and spotting common variables. Always underline or highlight common factors as you go.

Take the HCF outside the bracket. Divide each original term by the HCF — the quotients go inside the bracket. Double-check: number of terms inside bracket = number of original terms.

For grouping: Pair the terms so each pair has its own common factor. Factorise each pair, then look for the common binomial and pull it out.

Always verify by expanding your answer back. If you get the original expression, you are correct.

⛔ Board Exam Alert (Telangana, AP & CBSE): In factorisation questions, never leave a factorisable expression inside a bracket. Always check whether the terms inside can be factorised further. Stopping too early is the most common error and loses marks.

Common Mistakes to Avoid in Factorisation

Taking out only part of the HCF: In 6x² + 9xy, some students take out only 3 (ignoring the common variable x). The correct HCF is 3x, giving 3x(2x + 3y), not 3(2x² + 3xy).
Forgetting to write the remaining bracket: 6x² + 9xy = 3x(2x + 3y). Never write just "3x" — the bracket (2x + 3y) must always be there.
Wrong grouping in the grouping method: If your grouping doesn't lead to a common binomial after the first factorisation step, try pairing the terms differently.
Confusing prime factorisation of a number with factorisation of an expression: 36 = 2² × 3² is prime factorisation of a number. 3ab = 3 × a × b is factorisation of an algebraic term. The concept is the same; the notation differs.
Not verifying the answer: Always expand your factorised form and confirm it equals the original expression. This step takes 10 seconds and saves full marks.

Quick Reference — Key Concepts from This Introduction

Concept	Meaning	Example
Factor	A number/expression that divides another exactly	Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Prime Factorisation	Expressing a number as product of only prime numbers	36 = 2² × 3²
Irreducible Factor	A factor that cannot be split further	3ab → 3, a, b are irreducible
HCF of terms	Highest Common Factor — largest factor shared by all terms	HCF(6x², 9xy) = 3x
Common Factor Method	Take HCF outside; remainders go in bracket	6x² + 9xy = 3x(2x + 3y)
Grouping Method	Pair 4 terms → factorise pairs → pull common binomial	x²+xy+x+y = (x+y)(x+1)

What This Introduction Prepares You For

The two methods introduced here — the Common Factor Method and the Grouping Method — are the foundation for every exercise in Chapter 12. Exercise 12.1 gives extensive practice with the common factor technique, while Exercise 12.2 deepens your grouping skills. Later exercises cover factorisation using algebraic identities such as the difference of squares and perfect square trinomials — patterns you already know from Chapter 11 (Algebraic Identities).

In Class 9 Polynomials, factorisation is used to find the zeroes of polynomials and to solve division algorithms. In Class 10 Quadratic Equations, it is one of the primary methods for solving equations of the form ax² + bx + c = 0. Every factorisation skill you build here in Class 8 pays dividends throughout secondary school mathematics.

📐 Board Exam Tip (CBSE / Telangana / AP): Introduction-level concepts — irreducible factors, the need for factorisation, and the worked example (x² + 7x + 12) ÷ (x + 3) — often appear as 1-mark objective or fill-in-the-blank questions. Know the vocabulary: irreducible factor, prime factorisation, and HCF method.

✅ Chapter 12 at a Glance: Factorisation reverses multiplication. If expanding gives you a product → a sum, then factorisation takes a sum → back to a product. Master these two basic methods and the rest of Chapter 12 flows naturally.