Class 9 Mathematics Chapter 12 Factorisation: 18 Important Questions with Complete Solutions

The CBSE Class 9 Mathematics syllabus (rationalized 2024–25) places factorisation at the core of algebraic reasoning. Unlike rote memorization, factorisation demands conceptual understanding—students must identify the most efficient method for each expression, whether common factoring, grouping, or identity application. In recent board papers, 60–70% of algebra questions involve factorisation as a stepping stone to solving equations or simplifying fractions. Examiners test not just mechanical skill but problem-solving judgment: given an expression like 4x² + 12xy + 9y², can you instantly recognize the perfect square identity (2x + 3y)²? Or, when faced with 6xy + 9x + 4y + 6, can you regroup into (3x + 2)(2y + 3)? The question types we've curated reflect actual board expectations: MCQs test speed and identity recognition, 2-mark questions probe method selection, 3-mark questions combine multiple steps, and 5-mark long answers demand full working with logical flow. HOTS scenarios require you to apply factorisation to real contexts—fence areas, product dimensions, or polynomial division in practical settings. Mastering these 18 questions positions you to score 8/8 or 12/12 marks in the factorisation component of any Class 9 exam.

**Q1.** The factorization of 7x – 7 is: (a) 7(x – 1) (b) 7x(1) (c) x(7 – 7) (d) 7x – 1 **Answer: (a) 7(x – 1)** Explanation: 7 is the common factor in both terms. Extract it: 7x – 7 = 7(x – 1). **Q2.** Which of the following is a factor of 6a²b + 9ab²? (a) 3ab (b) 2a (c) 3b (d) 6ab **Answer: (a) 3ab** Explanation: 6a²b = 3ab × 2a and 9ab² = 3ab × 3b. Thus 3ab is the greatest common factor. **Q3.** Using the identity a² – b² = (a + b)(a – b), factorize 25x² – 16: (a) (5x – 4)² (b) (5x + 4)(5x – 4) (c) (25x – 16) (d) (5x)² – (4)² **Answer: (b) (5x + 4)(5x – 4)** Explanation: 25x² = (5x)² and 16 = 4². Apply the difference of squares: (5x)² – 4² = (5x + 4)(5x – 4). **Q4.** Factorize 4x² + 4xy + y²: (a) (2x + y)(2x + y) (b) (4x + y)² (c) 4(x + y)² (d) (2x – y)² **Answer: (a) (2x + y)(2x + y) or (2x + y)²** Explanation: This matches a² + 2ab + b² = (a + b)². Here a = 2x, b = y. So (2x + y)². **Q5.** When we divide 12a²b by 3ab, the quotient is: (a) 4ab (b) 4a (c) 12a (d) 4 **Answer: (b) 4a** Explanation: 12a²b ÷ 3ab = (12 ÷ 3) × (a² ÷ a) × (b ÷ b) = 4a × 1 = 4a.

**Q1.** Factorize: 3x + 6y + 9z **Solution:** The greatest common factor (GCF) of 3x, 6y, and 9z is 3. 3x + 6y + 9z = 3(x + 2y + 3z) **Verification:** 3(x) + 3(2y) + 3(3z) = 3x + 6y + 9z ✓ **Q2.** Factorize by regrouping: ax + ay + bx + by **Solution:** Group the terms in pairs: ax + ay + bx + by = (ax + ay) + (bx + by) Extract common factors from each group: = a(x + y) + b(x + y) Extract (x + y) as the common factor: = (x + y)(a + b) **Q3.** Using the identity a² + 2ab + b² = (a + b)², factorize: x² + 6x + 9 **Solution:** Recognize that x² = x², 9 = 3², and 6x = 2(x)(3). This fits the pattern a² + 2ab + b². So x² + 6x + 9 = (x + 3)² **Verification:** (x + 3)² = x² + 6x + 9 ✓ **Q4.** Divide 15m³n² by 5mn and simplify: **Solution:** 15m³n² ÷ 5mn = (15 ÷ 5) × (m³ ÷ m) × (n² ÷ n) = 3 × m² × n = 3m²n **Q5.** Factorize: 9x² – 25y² **Solution:** Recognize this as a² – b² where a = 3x and b = 5y. Using a² – b² = (a + b)(a – b): 9x² – 25y² = (3x)² – (5y)² = (3x + 5y)(3x – 5y) **Verification:** (3x + 5y)(3x – 5y) = 9x² – 15xy + 15xy – 25y² = 9x² – 25y² ✓

**Q1.** Factorize: 6xy + 4y + 9x + 6 **Solution:** Step 1: Group terms in pairs: 6xy + 4y + 9x + 6 = (6xy + 4y) + (9x + 6) Step 2: Extract common factor from each group: = 2y(3x + 2) + 3(3x + 2) Step 3: Extract (3x + 2) as the common factor: = (3x + 2)(2y + 3) **Verification:** (3x + 2)(2y + 3) = 6xy + 9x + 4y + 6 ✓ **Q2.** Factorize completely: 2x³ – 8x **Solution:** Step 1: Extract the greatest common factor (2x): 2x³ – 8x = 2x(x² – 4) Step 2: Recognize x² – 4 as a difference of squares (a² – b²): = 2x[(x)² – (2)²] Step 3: Apply a² – b² = (a + b)(a – b): = 2x(x + 2)(x – 2) **Verification:** 2x(x + 2)(x – 2) = 2x(x² – 4) = 2x³ – 8x ✓ **Q3.** Using identities, factorize: 4a² – 12ab + 9b² **Solution:** Step 1: Identify the pattern a² – 2ab + b²: 4a² = (2a)², 9b² = (3b)², and 12ab = 2(2a)(3b) Step 2: Apply a² – 2ab + b² = (a – b)²: 4a² – 12ab + 9b² = (2a)² – 2(2a)(3b) + (3b)² = (2a – 3b)² **Verification:** (2a – 3b)² = 4a² – 12ab + 9b² ✓ **Q4.** Divide the polynomial (6x² + 9x) by 3x and express as a factorization: **Solution:** Step 1: Factor out the common term 3x from the numerator: 6x² + 9x = 3x(2x + 3) Step 2: Divide by 3x: (6x² + 9x) ÷ 3x = [3x(2x + 3)] ÷ 3x = 2x + 3 Step 3: Write the factorization: 6x² + 9x = 3x(2x + 3) **Answer:** The quotient is 2x + 3, and the expression factors as 3x(2x + 3).

**Q1.** Factorize completely: 12x³y – 27xy³ **Complete Solution:** Step 1: Identify the greatest common factor (GCF). All terms contain 3, x, and y. 12x³y = 3xy × 4x² 27xy³ = 3xy × 9y² GCF = 3xy Step 2: Extract the common factor: 12x³y – 27xy³ = 3xy(4x² – 9y²) Step 3: Recognize 4x² – 9y² as a difference of squares: 4x² = (2x)² and 9y² = (3y)² Step 4: Apply a² – b² = (a + b)(a – b): 4x² – 9y² = (2x + 3y)(2x – 3y) Step 5: Write the complete factorization: 12x³y – 27xy³ = 3xy(2x + 3y)(2x – 3y) **Verification:** 3xy(2x + 3y)(2x – 3y) = 3xy[(2x)² – (3y)²] = 3xy(4x² – 9y²) = 12x³y – 27xy³ ✓ **Q2.** Factorize by regrouping: x³ + 2x² + 3x + 6 **Complete Solution:** Step 1: Group the first two and last two terms: x³ + 2x² + 3x + 6 = (x³ + 2x²) + (3x + 6) Step 2: Factor out the common factor from each group: From (x³ + 2x²): extract x² → x²(x + 2) From (3x + 6): extract 3 → 3(x + 2) Step 3: Rewrite: = x²(x + 2) + 3(x + 2) Step 4: Extract (x + 2) as the common factor: = (x + 2)(x² + 3) Step 5: Check if x² + 3 can be factored further. Since x² + 3 has no real factors, the final factorization is: x³ + 2x² + 3x + 6 = (x + 2)(x² + 3) **Verification:** (x + 2)(x² + 3) = x³ + 3x + 2x² + 6 = x³ + 2x² + 3x + 6 ✓ **Q3.** Use identities to factorize and then divide: (16a⁴ – 81b⁴) ÷ (4a² – 9b²) **Complete Solution:** Step 1: Factorize the numerator 16a⁴ – 81b⁴. Recognize this as a difference of squares where a = 4a² and b = 9b²: 16a⁴ = (4a²)² and 81b⁴ = (9b²)² 16a⁴ – 81b⁴ = (4a²)² – (9b²)² = (4a² + 9b²)(4a² – 9b²) Step 2: Further factorize 4a² – 9b² in the product: 4a² – 9b² = (2a)² – (3b)² = (2a + 3b)(2a – 3b) Step 3: Rewrite the numerator: 16a⁴ – 81b⁴ = (4a² + 9b²)(2a + 3b)(2a – 3b) Step 4: Divide by (4a² – 9b²): (16a⁴ – 81b⁴) ÷ (4a² – 9b²) = [(4a² + 9b²)(4a² – 9b²)] ÷ (4a² – 9b²) = 4a² + 9b² **Answer:** The quotient is 4a² + 9b².

**Case Study:** A rectangular garden has an area of (20x² + 50x) square meters. The length is (4x + 10) meters. A landscaper wants to factorize the area expression to find the width and then divide to verify. **Part (a):** Factorize the area expression 20x² + 50x. **Solution for (a):** Step 1: Identify the greatest common factor. 20x² = 10x × 2x 50x = 10x × 5 GCF = 10x Step 2: Extract the common factor: 20x² + 50x = 10x(2x + 5) **Answer:** The factorized form is 10x(2x + 5) square meters. **Part (b):** Find the width of the garden by dividing the area by the length. **Solution for (b):** Step 1: Write the width formula: Width = Area ÷ Length = (20x² + 50x) ÷ (4x + 10) Step 2: Factorize both numerator and denominator: Numerator: 20x² + 50x = 10x(2x + 5) [from part (a)] Denominator: 4x + 10 = 2(2x + 5) Step 3: Substitute: Width = [10x(2x + 5)] ÷ [2(2x + 5)] Step 4: Cancel the common factor (2x + 5): Width = (10x) ÷ 2 = 5x meters **Part (c):** Verify your answer by multiplying length × width. **Solution for (c):** Length × Width = (4x + 10) × 5x = 4x(5x) + 10(5x) = 20x² + 50x ✓ This matches the original area expression, confirming our answer. **Why This Matters:** Real-world geometry problems require both factorization and division skills. By recognizing common factors and simplifying, we solved for an unknown dimension without trial and error—a technique essential for advanced algebra and physics.

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