Factorisation — for Class 8

What Factorisation Actually Is

When we multiply $2 \times 3 \times 5 = 30$ , we call 2, 3, and 5 the factors of 30. Factorisation is the reverse — we start with 30 and break it apart into its factors.

In algebra, we do the same with expressions. If someone hands us $x^2 + 5x + 6$ , we ask: “What was multiplied to get this?” The answer is $(x + 2)(x + 3)$ . That process of breaking an expression into a product of simpler parts is factorisation.

Why does this matter for your exams? Two reasons. First, CBSE Class 8 has direct marks from factorisation questions — typically 4-6 marks in a section. Second, this is the foundation for everything in Class 9 (polynomials), Class 10 (quadratics), and eventually JEE algebra. Students who skip understanding the why here struggle badly in Class 10 when equations get harder.

The key mental shift: multiplication gives us one big expression, factorisation gives us a product of smaller ones. Both are the same mathematical object — just written differently.

Key Terms and Definitions

Factor: Any expression that divides another expression exactly, leaving no remainder. For $6x^2y$ , the factors include 2, 3, $x$ , $y$ , $x^2$ , $6x$ , $2x^2y$ , and so on.

Common Factor: A factor shared by every term in an expression. In $4x + 6y$ , the number 2 divides both 4 and 6 — so 2 is a common factor.

Highest Common Factor (HCF): The largest factor common to all terms. For $4x + 6y$ , HCF is 2. For $6x^2 + 9xy$ , HCF is $3x$ .

Irreducible Factor: A factor that cannot be broken down further. $(x + 3)$ is irreducible. $(x^2 - 9)$ is not irreducible — it factors into $(x+3)(x-3)$ .

Identity: An equation true for all values of the variable. The algebraic identities we use in factorisation are not something to prove — they are tools to recognise patterns.

Methods of Factorisation

Method 1: Taking Out the Common Factor

This is the first thing to check — always. Look at every term and pull out whatever divides all of them.

Step 1: Find the HCF of all terms (both numerical coefficients and variables).

Step 2: Write HCF outside a bracket and divide each term by the HCF inside.

Example: Factorise $12a^2b + 18ab^2$

Coefficients: HCF of 12 and 18 is 6
Variables: HCF of $a^2b$ and $ab^2$ is $ab$
So overall HCF = $6ab$

12a^2b + 18ab^2 = 6ab(2a + 3b)

Verification (do this always in exams): Expand back — $6ab \times 2a + 6ab \times 3b = 12a^2b + 18ab^2$ ✓

Always verify your factorisation by expanding. In CBSE, you get method marks, but an expansion check takes 15 seconds and confirms you haven’t made an error. Senior students do this automatically.

Method 2: Factorisation by Grouping

When there’s no single common factor across all terms, we group terms in pairs and find common factors within each pair.

Step 1: Group the terms into pairs (usually 2+2 for four-term expressions).

Step 2: Factorise each pair separately.

Step 3: If the bracket remaining in both pairs is the same, take that as a common factor.

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

ax + bx + ay + by

= x(a + b) + y(a + b)

= (a + b)(x + y)

The bracket $(a+b)$ appeared twice — that’s the clue grouping worked.

A very common error: students group incorrectly. If $x(a+b) + y(a-b)$ appears — brackets are different — regrouping is needed. Try $(ax + ay) + (bx + by)$ instead. Always check that the bracket is identical in both groups.

Trickier Example: Factorise $6xy - 4y + 6 - 9x$

Here, let’s group as $(6xy - 9x) + (6 - 4y)$ :

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

Hmm — brackets are $(2y-3)$ and $(3-2y)$ . These look different but $(3-2y) = -(2y-3)$ .

= 3x(2y - 3) - 2(2y - 3) = (2y - 3)(3x - 2)

Method 3: Using Algebraic Identities

Three identities dominate Class 8 factorisation. Recognising the pattern is the skill.

Identity 1 (Perfect Square — Sum):

a^2 + 2ab + b^2 = (a + b)^2

Identity 2 (Perfect Square — Difference):

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

Identity 3 (Difference of Squares):

a^2 - b^2 = (a + b)(a - b)

Pattern recognition is everything here. Ask yourself three questions when you see a quadratic-looking expression:

Is the first term a perfect square? Is the last term a perfect square?
Is the middle term twice the product of their square roots?
Is the sign of the last term positive (use Identity 1 or 2) or negative (check Identity 3 with no middle term)?

Example (Identity 1): Factorise $9x^2 + 12xy + 4y^2$

$9x^2 = (3x)^2$ ✓
$4y^2 = (2y)^2$ ✓
Middle term: $2 \times 3x \times 2y = 12xy$ ✓

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

Example (Identity 3): Factorise $49m^2 - 64n^2$

$49m^2 = (7m)^2$ , $64n^2 = (8n)^2$
No middle term — classic difference of squares

49m^2 - 64n^2 = (7m + 8n)(7m - 8n)

Solved Examples

Easy Level (CBSE Board Pattern)

Q1. Factorise $15pq + 20qr + 25rp$

HCF of 15, 20, 25 is 5. No common variable across all terms.

= 5(3pq + 4qr + 5rp)

Q2. Factorise $x^2 - 6x + 9$

Check Identity 2: $x^2 = (x)^2$ , $9 = (3)^2$ , middle term $= 2 \times x \times 3 = 6x$ ✓ (with negative sign)

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

Medium Level (CBSE + ICSE)

Q3. Factorise $4p^2 - 9q^2 + 6pr - 9qr$

Two terms look like difference of squares, two need grouping. Regroup:

(4p^2 - 9q^2) + (6pr - 9qr)

= (2p + 3q)(2p - 3q) + 3r(2p - 3q)

= (2p - 3q)(2p + 3q + 3r)

This type — combining Identity 3 with grouping — appears regularly in CBSE Class 8 unit tests and ICSE papers. The trick is to factor the “difference of squares” chunk first, then look for the matching bracket in the remaining terms.

Q4. Factorise $a^4 - b^4$

Recognise $a^4 = (a^2)^2$ and $b^4 = (b^2)^2$ — difference of squares!

a^4 - b^4 = (a^2 + b^2)(a^2 - b^2) = (a^2 + b^2)(a + b)(a - b)

Always apply identity again if any factor can be factorised further.

Hard Level (Olympiad / ICSE Extension)

Q5. Factorise $(x + y)^2 - (x - y)^2$

Let $A = (x+y)$ and $B = (x-y)$ . This is $A^2 - B^2 = (A+B)(A-B)$ .

= \big[(x+y) + (x-y)\big]\big[(x+y) - (x-y)\big]

= (2x)(2y) = 4xy

This approach — substituting to see the hidden pattern — is a powerful technique for olympiad-level questions.

Exam-Specific Tips

CBSE Class 8 (Board / Unit Tests)

CBSE marks factorisation for both method and answer. A typical 3-mark question expects:

1 mark: identifying the method / HCF
1 mark: intermediate step (correct bracket)
1 mark: final factorised form

Always write intermediate steps — never jump from the question to the answer.

CBSE internal assessment and unit tests commonly ask: “Factorise using appropriate identity.” This phrasing means they want you to name the identity used — write “Using Identity $a^2 - b^2 = (a+b)(a-b)$ ” before applying it. It earns you the method mark even if you make a small arithmetic error later.

ICSE Class 8

ICSE tends to give slightly more complex grouping problems and multi-step factorisation (like Q4 and Q5 above). Also expects candidates to state “complete factorisation” — meaning no factor can be further factorised.

Olympiads (NSO, IMO)

Olympiad questions often hide factorisation inside computation — for example, quickly calculating $99^2 - 1^2 = (99+1)(99-1) = 100 \times 98 = 9800$ . Train yourself to see the identity pattern in numbers, not just algebra.

Common Mistakes to Avoid

Mistake 1 — Incomplete factorisation. Writing $a^4 - b^4 = (a^2 + b^2)(a^2 - b^2)$ and stopping. You must continue: $(a^2 - b^2) = (a+b)(a-b)$ . CBSE deducts marks for this.

Mistake 2 — Wrong sign in perfect square. Students write $x^2 - 6x + 9 = (x+3)^2$ . Remember: if the middle term is negative, the bracket is $(x - 3)^2$ . Expand to check — $(x+3)^2 = x^2 + 6x + 9$ , which is different.

Mistake 3 — Not taking out the HCF first. Before using any identity, always check for a common factor. $2x^2 - 8$ should first become $2(x^2 - 4)$ , then $2(x+2)(x-2)$ . Jumping to identity without extracting 2 loses a mark.

Mistake 4 — Grouping mismatches. In grouping, if the brackets don’t match after factorising each pair, most students panic and give up. Try regrouping in a different way. Sometimes you need to rearrange terms before grouping.

Mistake 5 — Treating $(a - b)^2$ as $a^2 - b^2$ . These are completely different. $(a-b)^2 = a^2 - 2ab + b^2$ . The difference of squares $a^2 - b^2$ has no middle term. Mixing these up is the single most common error in Class 8 factorisation.

Practice Questions

Q1. Factorise $5x^2 + 15xy$

HCF is $5x$ .

5x^2 + 15xy = 5x(x + 3y)

Q2. Factorise $16m^2 - 40mn + 25n^2$

$16m^2 = (4m)^2$ , $25n^2 = (5n)^2$ . Middle term: $2 \times 4m \times 5n = 40mn$ ✓ (negative sign).

Using Identity 2:

16m^2 - 40mn + 25n^2 = (4m - 5n)^2

Q3. Factorise $36x^2 - 49y^2$

$36x^2 = (6x)^2$ , $49y^2 = (7y)^2$ . Difference of squares.

36x^2 - 49y^2 = (6x + 7y)(6x - 7y)

Q4. Factorise $pq + qr + ps + rs$

Group as $(pq + qr) + (ps + rs)$ :

= q(p + r) + s(p + r) = (p + r)(q + s)

Q5. Factorise $x^2y - xy^2 + xz - yz$

Group as $(x^2y - xy^2) + (xz - yz)$ :

= xy(x - y) + z(x - y) = (x - y)(xy + z)

Q6. Factorise $81 - (2x - y)^2$

Write $81 = (9)^2$ . This is $9^2 - (2x-y)^2$ — difference of squares with $a = 9$ , $b = (2x-y)$ .

= (9 + 2x - y)(9 - 2x + y)

Q7. Factorise $12x^2 - 27$

Always extract HCF first: HCF is 3.

12x^2 - 27 = 3(4x^2 - 9) = 3(2x + 3)(2x - 3)

Students who skip the HCF step get $\frac{1}{2}$ marks at best.

Q8. Factorise $x^4 - 16$

x^4 - 16 = (x^2)^2 - (4)^2 = (x^2 + 4)(x^2 - 4)

Now $(x^2 - 4) = (x+2)(x-2)$ . Final answer:

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

Note: $(x^2 + 4)$ cannot be factorised further (sum of squares, no real factors).

Q9. If $x + \frac{1}{x} = 5$ , find $x^2 + \frac{1}{x^2}$

Square both sides: $\left(x + \frac{1}{x}\right)^2 = 25$

x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 25

x^2 + 2 + \frac{1}{x^2} = 25

x^2 + \frac{1}{x^2} = 23

This type appears in ICSE Class 8 and uses Identity 1 in reverse.

Q10. Factorise $a^2 - b^2 + 2bc - c^2$

Rewrite the last three terms: $-b^2 + 2bc - c^2 = -(b^2 - 2bc + c^2) = -(b-c)^2$

So the expression becomes: $a^2 - (b-c)^2$

Using difference of squares:

= \big[a + (b-c)\big]\big[a - (b-c)\big] = (a + b - c)(a - b + c)

This is a Class 9-level question but good Class 8 students should be able to handle it.

FAQs

What is the difference between factors and multiples?

Factors are what you multiply together to get a number. Multiples are what you get when you multiply a number by integers. Factors of 12 are 1, 2, 3, 4, 6, 12. Multiples of 12 are 12, 24, 36, … They go in opposite directions.

Is factorisation the same as simplification?

Not exactly. Simplification is a broad term — it includes cancelling, combining like terms, etc. Factorisation specifically means writing as a product. All factorisations simplify an expression, but not all simplifications are factorisations.

How do I know which method to use?

Check in this order: (1) Is there a common factor? Always extract it first. (2) Is the expression a difference of two squares? (3) Does it match the perfect square pattern? (4) Can I group the terms? Working through this checklist every time builds the habit.

Why can’t we factorise $a^2 + b^2$ ?

The sum of squares has no real factors. You can verify: there’s no combination $(a + \text{something})(a - \text{something})$ that gives $a^2 + b^2$ without a middle term. This is why only difference of squares appears in Class 8.

What does “completely factorised” mean?

Every factor is irreducible — it cannot be broken down further. $2(x^2 - 4)$ is not completely factorised because $(x^2-4) = (x+2)(x-2)$ . The complete form is $2(x+2)(x-2)$ .

How is factorisation used in real life?

Whenever we simplify fractions, solve equations, or find LCM/HCF in higher classes, factorisation is the tool. In Class 10, you’ll solve quadratic equations by factorisation. In Class 11 and JEE, partial fractions use factorisation. It’s a foundational skill, not a standalone chapter.

Does factorisation appear in CBSE Class 8 final exam?

Yes — typically one 2-mark question (identify and apply an identity) and one 3-mark question (multi-step or grouping). Together, around 5 marks directly from this chapter, plus indirect application in fractions (rational expressions).

What if no method seems to work?

Re-examine whether you’ve extracted the HCF first. Most failed factorisations are caused by missing a common factor at the start. After extracting it, the remaining expression usually fits one of the patterns cleanly.