Factorisation Class 9 OP Malhotra Exe-4B ICSE Maths Solutions Ch-4. We Provide Step by Step Solutions / Answer of Questions on Factorisation OP Malhotra Maths. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics
Factorisation Class 9 OP Malhotra Exe-4B ICSE Maths Solutions Ch-4
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 9th |
| Chapter-4 | Factorisation |
| Writer | OP Malhotra |
| Exe-4B | Factorisation by Regrouping |
| Edition | 2025-2026 |
Factorisation by Regrouping
Factorization by regrouping involves rearranging an algebraic expression, which are then extracted to simplify the expression. This method is particularly useful when expressions don’t have a straightforward common factor across all terms, but can be grouped to reveal common factors within subgroups
Step for Factorisation by Regrouping
- Identify and rearrange terms: Group terms that share a common factor.
- Factor out common factors: Within each group, identify and factor out the common factor.
- Look for a common binomial factor: If, after factoring out the common factors, the remaining expressions within parentheses are identical, this is your common binomial factor.
- Final factorization: Express the factored expression as the product of the common binomial factor and the remaining expression
Exercise- 4B
Factorisation Class 9 OP Malhotra Exe-4B ICSE Maths Solutions Ch-4
Que-1: bx + 2b + cx + 2c
Sol: bx + 2b + cx + 2c
= bx + cx + 2b + 2c
= x(b + c) + 2 (b + c)
= (b + c) (x + 2)
Que-2: y² + 2y² + 3y + 6
Sol: y² + 2y² + 3y + 6
= y²(y + 2) + 3(y + 2)
= (y + 2) (y² + 3)
Que-3: xa + 3b + xb + 3a
Sol: xa + 3b + xb + 3a
= xz + xb + 3a + 3b
= x (a + b) + 3 (a + b)
= (a + b) (x + 3)
Que-4: 8xy + 5zy – 8xt – 5zt
Sol: 8xy + 5zy – 8xt – 5zt
= 8xy – 8xt + 5zy – 5zt
= 8x (y – t) + 5z (y – t)
= (y – r) (8x + 5z)
Que-5: 8kl + 12ml – 12mn – 8kn
Sol: 8kl + 12ml – 2mn – 8kn
= 8kl – 8kn + 12ml – 12mn
= 8k (l – n) + 12m (l – n)
= (l – n) (8k + 12m)
Que-6: 32 (x + y)² – 2x – 2y
Sol: 32 (x + y)² – 2x – 2y
= 32 (x + y) (x + y) – 2 (x + y)
= 2(x + y)[16(x + y) – 1]
= 2 (x + y) (16x + 16y – 1)
Que-7: x³ – x² + ax + x – a – 1
Sol: x³ – x² + ax + x – a – 1
= x³ – x² + ax – a + x – 1
= x² (x – 1) + a (x – 1) + 1 (x – 1)
= (x – 1) (x² + a + 1)
Que-8: ab (c² + 1) + c (a² + b²)
Sol: ab (c² + 1) + c (a² + b²)
= abc² + ab + a²c + b²c
= abc² + a²c + b²c + ab
= ac (be + a) + b (bc + a)
= (bc + a) (ac + b)
Que-9: a³ – a² + xa + a – x – 1
Sol: a³ – a² + xa + a – x – 1
= a³ – a² + xa – x + a – 1
= a² (a – 1) + x (a – 1) + 1 (a – 1)
= (a – 1) (a² + x + 1)
Que-10: 6a³b + 3a²b² – 2a²b – ab²
Sol: 6a³b + 3a²b² – 2a²b – ab²
= 3a²b (2a + b)- ab (2a + b)
= (2a + b) (3 a²b – ab)
= (2a + b) ab (3a – 1)
= ab (3a – 1) (2a + b)
Que-11: a³ + ab (1 – 2a) – 2b²
Sol: a³ + ab (1 – 2d) – 2b²
= a³ + ab – 2 a²b – 2 b²
= a (a² + b) – 2b (a² + b)
= (a² + b) (a- 2b)
Que-12: (p² + 1)q – p² – q²
Sol: (p² + 1 )q – p² – q²
= p²q + q – p² – q²
= p²q – p² – q² + q
=p² (q – 1) – q(q – 1)
= (q – 1) (p² – q).
— : End of Factorisation Class 9 OP Malhotra Exe-4B ICSE Maths step by step Solutions Ch-4 :–
Return to :- OP Malhotra Class 9 ICSE Maths Solutions
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