Factorisation Class 9 OP Malhotra Exe-4F ICSE Maths Solutions Ch-4. We Provide Step by Step Solutions / Answer of Questions on Factorisation by converting into Difference of two squires. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.
Factorisation Class 9 OP Malhotra Exe-4F ICSE Maths Solutions
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 9th |
| Chapter-4 | Factorisation |
| Writer | OP Malhotra |
| Exe-4F | Factorisation by splitting middle term |
| Edition | 2025-2026 |
Exercise- 4F
Factorisation Class 9 OP Malhotra Exe-4F ICSE Maths Solutions Ch-4.
Factorise :
Que-1: a² + 5a + 6
Sol: a² + 5a + 6
= a² + 2a + 3a + 6
= a (a + 2) + 3 (a + 2)
= (a + 2) (a + 3)
Que-2: a² + 6a + 8
Sol: a² + 6a + 8
= a² + 2 a + 4a + 8
= a (a + 2) + 4 (a + 2)
= (a + 2) (a + 4)
Que-3: p² + 10p + 16
Sol: p² + 10p + 16
= p² + 2p + 8p + 16
= p(p + 2) + 8(p + 2)
= (p + 2)(p + 8)
Que-4: a² + 13a + 42
Sol: a² + 13a + 42
= a² + 7a + 6a + 42
= a (a + 7) + 6 (a + 7)
= (a + 7) (a + 6)
Que-5: a² + 25a – 54
Sol: a² + 25a – 54
= a (a + 27) – 2 (a + 27)
= (a + 27) (a – 2)
Que-6: x² + 5x – 176
Sol: x² + 5x – 176
= x² + 16a – 11a – 176
= x (x + 16) – 11 (a + 16)
= (x + 16) (x – 11)
Que-7: y² – 18y + 65
Sol: y² – 18y + 65
= y² – 13y – 5y + 65
= y (y – 13) – 5 (y – 13)
= (y – 13) (y – 5)
Que-8: m² – 29m + 204
Sol: m² – 29m + 204
= m² – 17a – 12m + 204
= m (m – 17) – 12 (m – 17)
= (m – 17) (m – 12)
Que-9: b² – 2b – 48
Sol: b² – 2b – 48
= b² – 8b + 6b – 48
= b (b – 8) + 6 (b – 8)
= (b – 8) (b + 6)
Que-10: x² – 11x – 102
Sol: x² – 11x – 102
= x² – 17x + 6x – 102
= x (x – 17) + 6 (x – 17)
= (x – 17) (x + 6)
Que-11: 3 – 4t + t²
Sol: 3 – 4t + t²
= 3 – 3t – t + t²
= 3(1 – t) – t(1 – t)
= (1 – t) (3 – t)
Que-12: 51 – 20k + k²
Sol: 51 – 20k + k²
= 51 – 17k – 3k + k²
= 17 (3 – k) – k (3 – k)
= (3 – k) (17 – k)
Que-13: 2x² – 10x + 12
Sol: 2x² – 10x + 12
= 2 (x² – 5x + 6)
= 2 [x² – 2x – 3x + 6]
= 2 [x (x – 2) – 3 (x – 2)]
= 2 (x – 2) (x – 3)
Que-14: 3x³ – 33x² + 84x
Sol: 3x³ – 33x² + 84x
= 3x [x² – 11x + 28]
= 3x [x² – 7x – 4x + 28]
= 3x [x (x – 7) – 4 (x – 7)]
= 3x (x – 7) (x – 4)
Que-15: 5y² – 45y – 110
Sol: 5y² – 45y – 110 = 5 (y² – 9y – 22)
= 5 [y² – 11y + 2y – 22]
= 5 [y (y – 11) + 2(y – 11)]
= 5 (y – 11) (y + 2)
Que-16: x4 – 13x² + 36
Sol: x4 – 13x² + 36
= x4 – 9x² – 4x² + 36
= x² (x² – 9) – 4 (x² – 9)
= (x² – 9) (x² – 4)
= [(x)² – (3)²] [(x)² – (2)²]
= (x + 3) (x – 3) (x + 2) (x – 2)
Que-17: x² + 3xy – 88y²
Sol: x² + 3xy – 88y²
= x² + 11 xy – 8xy – 88y²
= x (x + 11y) – 8y (x + 11y)
= (x + 11 y) (x – 8y)
Que-18: x4 – x²y² – 72y4
Sol: x4 – x²y² – 72y4
= x4 – 9x²y² + 8x² y² – 72y4
= x² (x² – 9y²) + 8y² (x² – 9y²)
= (x² – 9y²) (x² + 8y²)
= (x² + 8y²) [(x)² – (3y)²]
= (x² + 8y) (x + 3y) (x – 3y)
= (x + 3y) (x – 3y) (x² + 8y²)
Que-19: a³b³ – 9a²b² + 20ab
Sol: a³b³ – 9a²b² + 20ab
= ab [a²b² – 9ab + 20]
= ab [a²b² – 4ab – 5ab + 20]
= ab [ab (ab – 4) – 5 (ab – 4)]
= ab (ab – 4) (ab – 5)
Que-20: (x² + x)² + 4 (x² + x) – 21
Sol: (x² + x)² + 4 (x² + x) – 21
Let x² + x = a, then
a² + 4a – 21 = a² + 7a – 3a – 21
= a (a + 7) – 3 (a – 7)
= (a + 7) (a – 3)
= (x² + x + 7) (x² + x – 3)
— : End of Factorisation Class 9 OP Malhotra Exe-4F ICSE Maths step by step Solutions Ch-4 :–
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