Factorisation Class 9 OP Malhotra Exe-4G ICSE Maths Solutions Ch-4 We Provide Step by Step Solutions / Answer of Questions on Factorisation by splitting middle term. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.
Factorisation Class 9 OP Malhotra Exe-4G ICSE Maths Solutions
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 9th |
| Chapter-4 | Factorisation |
| Writer | OP Malhotra |
| Exe-4G | Factorisation by splitting middle term |
| Edition | 2025-2026 |
Exercise- 4G
Factorisation Class 9 OP Malhotra Exe-4G ICSE Maths Solutions Ch-4
Que-1: 2x² + 3x + 1
Sol: 2x² + 3x + 1
= 2x² + 2x + x + 1
= 2x (x + 1) + 1 (x + 1)
= (x + 1) (2x + 1)
Que-2: 3x² + 17x + 10
Sol: 3x²+17x+10
= 3x² + 15x + 2x + 10
= 3x (x + 5) + 2 (x + 5)
= (x + 5) (3x + 2)
Que-3: 5x² + 9x – 2
Sol: 5x² + 9x – 2
= 5x² + 10x – x – 2
= 5x (x + 2) – 1 (x + 2)
= (x + 2) (5x – 1)
Que-4: 6x² – 7x – 5
Sol: 6x² – 7x – 5
= 6x² – 10x + 3x – 5
= 2x (3x – 5) + 1 (3x – 5)
= (3x – 5) (2x + 1)
Que-5: 6y² – 17y + 12
Sol: 6y² – 17y + 12
= 6y² – 9y – 8y + 12
= 3y – (2y – 3) – 4 (2y – 3)
= (2y – 3) (3y – 4)
Que-6: 8y² – 2y – 1
Sol: 8y² – 2y – 1
= 8y² – 4y + 2y – 1
= 4y (2y – 1) + 1 (2y – 1)
= (2y – 1) (4y + 1)
Que-7: 18bx² + 18bx – 20b
Sol: 18bx² + 18bx – 20b
= 2b (9x² + 9x – 10)
= 2b (9x² + 15x – 6x – 10)
= 2b [3x (3x + 5) – 2 (3x + 5)]
= 2b (3x + 5) (3x – 2)
Que-8: 14x² – 60xy + 16y²
Sol: 14x² – 60xy + 16y²
= 2 [7x² – 30xy + 8y²]
= 2 [7x² – 28xy – 2xy + 8y²]
= 2[7x(x – 4y) – 2y(x – 4y)]
= 2 (x – 4y) (7x – 2y)
Que-9: 30x² + 103xy – 7y²
Sol: 30x² + 103xy – 7y²
= 30x² + 105xy – 2xy – 7y²
= 15x (2x + 7y) – y (2x + 7y)
= (2x + 7y) (15x – y)
Que-10: 12x² – 29xy + 14y²
Sol: 12x² – 29xy + 14y²
= 12x² – 8xy – 21xy + 14y²
= 4x (3x – 2y) – 7y (3x – 2y)
= (3x – 2y) (4x – 7y)
Que-11: 15 + p (7 – 2p)
Sol: 15 + p (7 – 2p) = 15 + 7p – 2p²
= 15 + 10p – 3p – 2p²
= 5 (3 + 2p) – p (3 + 2p)
= (3 + 2p) (5 – p)
Que-12: 2 – y(7 – 5y)
Sol: 2 – y(7 – 5y)
= 2 – 7y + 5y²
= 2 – 5y – 2y + 5y²
= 1 (2 – 5y) – y(2 – 5y)
= (2 – 5y) (1 – y)
Que-13: x(2x + 5) – 3
Sol: x (2x + 5) – 3
= 2x² + 5x – 3
= 2x² + 6x – x – 3
= 2x (x + 3) – 1 (x + 3)
= (x + 3) (2x – 1)
Que-14: 1 – 18y – 63y²
Sol: 1 – 18y – 63y²
= 1 – 21y + 3y – 63y²
= 1 (1 – 21y) + 3y(1 – 21y)
= (1 – 21y) (1 + 3y)
Que-15: 8a³b – 10a²b² – 12ab³
Sol: 8a³b – 10a³b³ – 12ab³
= 2ab (4a² – 5ab – 6b²)
= 2ab [4a² – 8ab + 3ab – 6b²]
= 2ab [4a (a – 2b) + 3 b (a – 2b)]
= 2ab (a – 2b) (4a + 3b)
Que-16: 2 (a + b)² – 5 (a + b) – 3
Sol: 2 (a + b)² – 5 (a + b) – 3
Let a + b = x, then
2x² – 5x – 3 = 2x² – 6x + x – 3
= 2x (x – 3) + 1 (x – 3)
= (x – 3) (2x + 1)
(Substituting the value of x)
= (a + b – 3) (2a + 2b + 1).
— : End of Factorisation Class 9 OP Malhotra Exe-4G ICSE Maths Solutions Ch-4 :–
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