Exercise 12.1 — Common Factors

Factorisation method and grouping the terms.

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What Is Factorisation?

Factorisation (also spelled factorization) is the process of writing an algebraic expression as a product of its factors. Think of it as the reverse of multiplication: instead of expanding brackets, you compress an expression back into a multiplied form.

For example, the number 12 can be written as 2 × 6, or 3 × 4, or 2 × 2 × 3. Similarly, the expression 6x + 9 can be written as 3(2x + 3). Factorisation is one of the most powerful tools in algebra — it simplifies expressions, helps solve equations, and appears throughout Class 8, 9, and 10 mathematics in CBSE, Telangana, and Andhra Pradesh board examinations.

Step 1Find the HCF of all terms

Step 2Take the HCF outside

Step 3Write remaining terms inside brackets

Step 4Verify by expanding back

8x² + 12xy ⬇ 4x (2x + 3y) HCF = 4x → taken outside the bracket

💡 Golden rule: To verify your factorisation is correct, always expand the bracket mentally. If you get back the original expression, your answer is right.

Question 1 — Finding Common Factors of Given Terms

Before you can factorise an expression, you need to find the common factors of its individual terms. This means listing all factors of each term, identifying those that appear in all of them, and listing every such common factor — not just the Highest Common Factor (HCF), but all common factors including 1.

The method used in Exercise 12.1 is prime factorisation: break each term into its prime building blocks (smallest prime numbers and individual variables), then read off the common pieces.


  To find all common factors:

  1. Write each term as a product of primes and variables

  2. Find HCF = product of primes/variables common to ALL terms

  3. List ALL factors of the HCF (those are the common factors)

Q1 (i)

Find the common factors of 8x and 24

Prime Factorisation

8x = 2 × 2 × 2 × x

24 = 2 × 2 × 2 × 3

Common prime factors: 2 × 2 × 2 HCF = 8

Common factors of 8x and 24 → 1, 2, 4, 8

📌 Note: The variable x appears in 8x but not in 24, so x is NOT a common factor. Only the numerical part (8) is shared.

Q1 (ii)

Find the common factors of 3a and 21ab

3a = 3 × a

21ab

21ab = 3 × 7 × a × b

Common prime factors: 3 and a HCF = 3a

Common factors → 1, 3, a, 3a

📌 Notice that 'b' appears only in 21ab, not in 3a. So 'b' is not a common factor. The variable 'a' is shared, so it is included in the HCF.

Q1 (iii)

Find the common factors of 7xy and 35x²y³

7xy

7xy = 7 × x × y

35x²y³

35x²y³ = 7 × 5 × x × x × y × y × y

Common factors: 7, x, and y (lowest powers) HCF = 7xy

Common factors → 1, 7, x, y, 7x, 7y, xy, 7xy

💡 When variables appear in both terms, use the lower power for the HCF. Here x appears as x¹ and x², so we take x¹ = x. Similarly y appears as y¹ and y³, so we take y¹.

Q1 (iv)

Find the common factors of 4m², 6m², and 8m³

4m²

4m² = 2 × 2 × m × m

6m²

6m² = 2 × 3 × m × m

8m³

8m³ = 2 × 2 × 2 × m × m × m

HCF

2 × m × m = 2m²

Common factors → 1, 2, m, 2m, m², 2m²

📌 With three terms, find the HCF of all three together. The factor 2 appears in all three (4m² has 2², 6m² has 2¹, 8m³ has 2³) — so we take the minimum power: 2¹ = 2. Similarly m appears as m², m², m³ → minimum is m².

Q1 (v)

Find the common factors of 15p, 20qr, and 25rp

15p

15p = 3 × 5 × p

20qr

20qr = 2 × 2 × 5 × q × r

25rp

25rp = 5 × 5 × r × p

HCF

Common factors → 1, 5

📌 Only the number 5 is shared across all three terms. No variable (p, q, r) appears in every single term — so no variable makes it into the HCF. This is a reminder to always check all terms before including a factor.

Q1 (vi)

Find the common factors of 4x², 6xy, and 8y²x

4x²

4x² = 2 × 2 × x × x

6xy

6xy = 2 × 3 × x × y

8y²x

8y²x = 2 × 2 × 2 × y × y × x

HCF

2 × x = 2x

Common factors → 1, 2, x, 2x

📌 The variable y appears in 6xy and 8y²x but NOT in 4x², so y is excluded from the HCF.

Q1 (vii)

Find the common factors of 12x²y and 18xy²

12x²y

12x²y = 2 × 2 × 3 × x × x × y

18xy²

18xy² = 2 × 3 × 3 × x × y × y

Common factors: 2, 3, x, y → HCF = 2 × 3 × x × y HCF = 6xy

Common factors → 1, 2, 3, 6, x, y, 2x, 2y, 3x, 3y, xy, 6x, 6y, 2xy, 3xy, 6xy

✅ This is the most comprehensive list in Q1 because 6xy has many sub-factors (1, 2, 3, 6 numerically and x, y, xy for variables — and all their combinations).

Question 2 — Factorise the Expressions (Taking Out Common Factor)

In Question 2, you move from identifying common factors to using them to rewrite an expression in factorised form. The process is: find the HCF of all terms → take it outside the bracket → divide each term by the HCF and write the quotients inside the bracket.


  If HCF of (ax² + bxy) = x, then:

  ax² + bxy = x(ax + by)

Q2 (i)

Factorise: 5x² − 25xy

5x²

5x² = 5 × x × x

25xy

25xy = 5 × 5 × x × y

HCF of 5x² and 25xy = 5 × x = 5x

5x² − 25xy = 5x(x) − 5x(5y) = 5x(x − 5y)

✅ Check: 5x × x = 5x² ✓ and 5x × 5y = 25xy ✓

Q2 (ii)

Factorise: 9a² − 6ax

9a²

9a² = 3 × 3 × a × a

6ax

6ax = 2 × 3 × a × x

HCF = 3 × a = 3a

9a² − 6ax = 3a(3a) − 3a(2x) = 3a(3a − 2x)

Q2 (iii)

Factorise: 7p² + 49pq

7p²

7p² = 7 × p × p

49pq

49pq = 7 × 7 × p × q

HCF = 7 × p = 7p

7p² + 49pq = 7p(p) + 7p(7q) = 7p(p + 7q)

⛔ Common mistake: Some students write the answer as 7p(p − 7q) because of the minus sign habit from the previous question. This is a PLUS expression — the sign stays positive.

Q2 (iv)

Factorise: 36a²b − 60a²bc

36a²b

= 2 × 2 × 3 × 3 × a × a × b

60a²bc

= 2 × 2 × 3 × 5 × a × a × b × c

HCF = 2 × 2 × 3 × a × a × b = 12a²b

36a²b − 60a²bc = 12a²b(3) − 12a²b(5c) = 12a²b(3 − 5c)

💡 The HCF here has three parts: a number (12), a squared variable (a²), and a single variable (b). Always identify all three types of common elements.

Q2 (v)

Factorise: 3a²bc + 6ab²c + 9abc²

3a²bc = 3 × a × a × b × c 6ab²c = 2 × 3 × a × b × b × c 9abc² = 3 × 3 × a × b × c × c

HCF = 3 × a × b × c = 3abc

= 3abc(a) + 3abc(2b) + 3abc(3c) = 3abc(a + 2b + 3c)

✅ Three-term factorisation works exactly the same way — find HCF of all three terms, take it out, and write the three quotients inside one bracket.

Q2 (vi)

Factorise: 4p² + 5pq − 6pq²

4p² = 2 × 2 × p × p 5pq = 5 × p × q 6pq² = 2 × 3 × p × q × q

HCF = p (only p is common to all three)

= p(4p) + p(5q) − p(6q²) = p(4p + 5q − 6q²)

📌 When the HCF is just a single variable (p), the factorisation still works the same — divide every term by p and place them inside the bracket.

Q2 (vii)

Factorise: ut + at²

ut = u × t at² = a × t × t

HCF = t

= t(u) + t(at) = t(u + at)

💡 This expression relates to the physics equation of motion: s = ut + ½at². Factorising t from both terms is the first step in simplifying that formula — real-world maths!

Question 3 — Factorisation by Regrouping Terms

Sometimes, no single factor is common to all terms of an expression. In such cases, we use a powerful technique called factorisation by grouping: split the expression into two pairs, factor out the common factor from each pair separately, and then notice that a common bracket appears — which you then take out as the final factor.


  ax + ay + bx + by

  = a(x + y) + b(x + y)   ← same bracket appears

  = (a + b)(x + y)

💡 Key insight for grouping: After factoring each pair, both groups must produce the same bracket. If they don't, try rearranging the terms and grouping differently.

Q3 (i)

Factorise: 3ax − 6xy + 8by − 4ab

1 Group into two pairs: = (3ax − 6xy) + (8by − 4ab)

2 Factor each pair: = 3x(a − 2y) + 4b(2y − a)

3 Rewrite second group to match bracket: = 3x(a − 2y) − 4b(a − 2y)

4 Take out common bracket (a − 2y): = (3x − 4b)(a − 2y)

⛔ Sign alert: The second pair gives 4b(2y − a). To match the bracket (a − 2y), flip the sign inside and write −4b(a − 2y). This is the trickiest step — missing this sign flip is the most common error in regrouping questions.

Q3 (ii)

Factorise: x³ + 2x² + 5x + 10

1 Group into two pairs: = (x³ + 2x²) + (5x + 10)

2 Factor each pair: = x²(x + 2) + 5(x + 2)

3 Both pairs have (x + 2) — take it out: = (x² + 5)(x + 2)

✅ Clean regrouping — both pairs naturally give the same bracket (x + 2). This is the most straightforward grouping case.

Q3 (iii)

Factorise: m² − mn + 4m − 4n

1 Group: = (m² − mn) + (4m − 4n)

2 Factor each pair: = m(m − n) + 4(m − n)

3 Common bracket (m − n): = (m + 4)(m − n)

Q3 (iv)

Factorise: a³ − a²b² − ab + b³

1 Group: = (a³ − a²b²) − (ab − b³)

2 Factor each group: = a²(a − b²) − b(a − b²)

3 Common bracket (a − b²): = (a² − b)(a − b²)

📌 Here the grouping uses subtraction between the two pairs. The second group −(ab − b³) = −b(a − b²), which matches the first bracket. This requires careful sign handling.

Q3 (v)

Factorise: p²q − pr² − pq + r²

1 Group: = (p²q − pr²) − (pq − r²)

2 Factor each group: = p(pq − r²) − 1(pq − r²)

3 Common bracket (pq − r²): = (p − 1)(pq − r²)

💡 When a group like (pq − r²) has no numerical factor, you still factor out 1 to make the pattern visible. This is a valid step.

Quick Reference — All Answers at a Glance

Question	Expression	HCF / Method	Factorised Form
Q1(i)	8x, 24	HCF = 8	1, 2, 4, 8
Q1(ii)	3a, 21ab	HCF = 3a	1, 3, a, 3a
Q1(iii)	7xy, 35x²y³	HCF = 7xy	1, 7, x, y, 7x, 7y, xy, 7xy
Q1(iv)	4m², 6m², 8m³	HCF = 2m²	1, 2, m, 2m, m², 2m²
Q1(v)	15p, 20qr, 25rp	HCF = 5	1, 5
Q1(vi)	4x², 6xy, 8y²x	HCF = 2x	1, 2, x, 2x
Q1(vii)	12x²y, 18xy²	HCF = 6xy	1, 2, 3, 6, x, y, 6x, 6y, … 6xy
Q2(i)	5x² − 25xy	HCF = 5x	5x(x − 5y)
Q2(ii)	9a² − 6ax	HCF = 3a	3a(3a − 2x)
Q2(iii)	7p² + 49pq	HCF = 7p	7p(p + 7q)
Q2(iv)	36a²b − 60a²bc	HCF = 12a²b	12a²b(3 − 5c)
Q2(v)	3a²bc + 6ab²c + 9abc²	HCF = 3abc	3abc(a + 2b + 3c)
Q2(vi)	4p² + 5pq − 6pq²	HCF = p	p(4p + 5q − 6q²)
Q2(vii)	ut + at²	HCF = t	t(u + at)
Q3(i)	3ax − 6xy + 8by − 4ab	Grouping	(3x − 4b)(a − 2y)
Q3(ii)	x³ + 2x² + 5x + 10	Grouping	(x² + 5)(x + 2)
Q3(iii)	m² − mn + 4m − 4n	Grouping	(m + 4)(m − n)
Q3(iv)	a³ − a²b² − ab + b³	Grouping	(a² − b)(a − b²)
Q3(v)	p²q − pr² − pq + r²	Grouping	(p − 1)(pq − r²)

Common Mistakes to Avoid in Factorisation

Not using the full HCF: Students sometimes take out only a partial common factor (e.g., taking out 3 instead of 12a²b in Q2(iv)). Always fully factorise — find the complete HCF before taking it out.
Forgetting to list ALL common factors in Q1: The question asks for all common factors, not just the HCF. If the HCF is 6xy, you must also list 1, 2, 3, 6, x, y, 2x, 2y, 3x, 3y, xy, 2xy, 3xy, and 6xy.
Sign errors in regrouping (Q3): When the second pair yields a bracket like (2y − a), you must flip it to −(a − 2y) to match the first pair's bracket. Missing this flip gives the wrong answer.
Including variables not present in every term: In Q1(v) — 15p, 20qr, 25rp — students may include r or p in the HCF. But 20qr has no p, so p cannot be a common factor of all three terms.
Incorrect grouping: In Q3, if your two groups don't produce the same bracket, try regrouping the terms in a different order before concluding it can't be done.
Not verifying the answer: Always expand your factorised form mentally to check you get back the original expression — a step that takes 10 seconds and saves full marks.

⛔ Board exam alert (Telangana, AP, CBSE): In Q3 (regrouping), always show the intermediate step of factoring each pair separately — i.e., write 3x(a − 2y) − 4b(a − 2y) as a separate line before writing the final answer. Examiners award marks for method, not just the answer.

What Exercise 12.1 Prepares You For

The skills in this exercise are the foundation for the rest of Chapter 12 — Factorisation. The common-factor method learned in Q1 and Q2 is used directly in Exercise 12.2, where you use algebraic identities to factorise expressions like perfect squares and difference of squares.

The regrouping method from Q3 is essential for Exercise 12.3, which deals with factorising quadratic trinomials of the form x² + bx + c — a topic that bridges directly into Class 9 Polynomials.

For students in CBSE, Telangana, and Andhra Pradesh boards, factorisation appears in 2-mark, 3-mark, and 4-mark questions in SA-1 and SA-2 examinations. Strong command of HCF-based factorisation and regrouping is essential for scoring in Chapter 12. The regrouping technique also underpins simplification of rational expressions in Class 9 and 10.

📐 Study tip: Practice finding the HCF of algebraic terms quickly — this is the single most important micro-skill in this chapter. Try writing out the prime factorisation step mentally (without writing) to build speed for exams. For regrouping, always write all four steps explicitly in your exam answer sheet.