Factorisation Class 9 OP Malhotra Exe-4E 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-4E ICSE Maths Solutions Ch-4
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 9th |
| Chapter-4 | Factorisation |
| Writer | OP Malhotra |
| Exe-4E | Factorisation by converting into Difference of two squires |
| Edition | 2025-2026 |
Factorisation by Converting into Difference of Two Squires
- Identify a perfect square: Look for an expression that can be written as the square of a term (e.g., x², 4, 9y²).
- Identify another perfect square: Look for another perfect square that is being subtracted from the first perfect square.
- Rewrite the expression: Express the first term as a² and the second term as b².
- Apply the formula: Factor the expression using the formula a² – b² = (a + b)(a – b).
Exercise– 4E
Factorisation Class 9 OP Malhotra Exe-4E ICSE Maths Solutions Ch-4
Que-1: 7x² – 7
Sol: 7x² – 7
= 7 (x² – 1)
= 7[(x)² – (1)²]
= 7(x + 1)(x – 1) {∵ a² – b² = (a + b) (a – b)}
Que-2: 8 – 50y²z²
Sol: 8 – 50y²z²
= 2 (4 – 25y²z²)
= 2[(2)² – (5yz)²]
= 2 (2 + 5yz) (2 – 5yz) {∵ a² – b² = (a + b) (a – b)}
Que-3: ab² – ac²
Sol: ab² – ac²
= a (b² – c²)
= a [(b)² – (c)²]
= a (b + c) (b – c) {∵ a² – b² = (a + b) (a – b)}
Que-4: 36x³ – x
Sol: 36x³ – x
= x (36x² – 1)
= x[(6x)²-(l)²]
= x (6x + 1) (6x – 1) (∵ a² – b² = (a + b) (a – b)}
Que-5: x (x² – 1) + 7 (x² – 1)
Sol: x (x² – 1) + 7 (x² – 1)
= (x² – 1) (x + 7)
= [(x)² – (1)²](x + 7)
= (x + 1) (x – 1) (x + 7) {∵ a² – b² = (a + b) (a – b)}
Que-6: t² (t – 3)² – (t – 3)²
Sol: t² (t – 3)² – (t – 3)²
= (t – 3)² [t² – 1]
= (t – 3)² [(t)² – (1)²]
= (t – 3)² (t + 1) (t – 1) {∵ a² – b² = (a + b) (a – b)}
Que-7: 5c² (c + 2)² – 45 (c + 2)²
Sol: 5c² (c + 2)² – 45 (c + 2)²
= 5 (c + 2)² [c² – 9]
= 5 (c + 2)² [(c)² – (3)²]
= 5 (c + 2)² (c + 3) (c – 3) {∵ a² – b² = (a + b) (a – b)}
Que-8: (a + b)² – 1
Sol: (a + b)² – 1
= (a + b)² – (1)²
= (a + b + 1) (a + b – 1) {∵ a² – b² = (a + b) (a – b)}
Que-9: 1 – (x – y)²
Sol: 1 – (x – y)²
= (1)² – (x – y)²
= (1 + x – y) (1 – x + y) {∵ a² – b² = (a + b) (a – b)}
Que-10: 25a² – 16 (x – y)²
Sol: 25a² – 16 (x – y)²
= (5a)² – [4 (x – y)]²
= [5a + 4 (x – y)] [5a – 4 (x – y)] {∵ a² – b² = (a + b) (a – b)}
= (5a + 4x – 4y) (5a – 4x + 4y)
Que-11: 20 – 45 (m + n)²
Sol: 20 – 45 (m + n)²
= 5 [4 – 9 (m + n)²]
= 5 [(2)² – {3 (m + n)}²]
= 5 [2 + 3(m + n)] [2 – 3 (m + n)] {∵ a² – b² = (a + b) (a – b)}
= 5 [2 + 3m + 3 n] [2 – 3 m – 3 n]
Que-12: x4 – y4
Sol: x4 – y4
= (x²)² – (y²)²
= (x² + y²)(x² – y²) {∵ a² – b² = (a + b) (a – b)}
= (x² + y²) [(x)² – (y)²]
= (x² + y²)(x + y) (x – y)
= (x – y)(x + y)(x² + y²)
Que-13: x4 – 625
Sol: x4 – 625
= (x²)² – (25)²
= (x² + 25) (x² – 25) {∵ a² – b² = (a + b) (a – b)}
= (x² + 25) [(x)² – (5)²]
= (x² + 25) (x + 5) (x – 5)
= (x – 5) (x + 5) (x² + 25)
Que-14: xy5 – yx5
Sol: xy5 – yx5
= xy (y4 – x4)
= xy[(y²)²-(x²)²]
= xy(y² + x²) (y² – x²) {∵ a² – b² = (a + b) (a – b)}
= xy (y² + x²) [(y)² – (x)²]
= xj (y² + x²) (y + x) (y – x)
= xy (y – x) (y + x) (y² + x²)
Que-15: 81x4 – 256y4
Sol: 81x4 – 256y4
= (9x²)² – (16y²)²
= (9x² – 16y²) (9x² + 16y²) {∵ a² – b² = (a + b) (a – b)}
= [(3x)² – (4y)²] (9x² + 16y²)
= (3x – 4y) (3x + 4y) (9x² + 16y²)
Que-16: a² + ac + bc – b²
Sol: a² + ac + bc – b²
= a² – b² + ac + bc
= (a)² – (b)² + c (a + b)
= (a + b) (a – b) + c (a + b)
= (a + b) (a – b + c)
Que-17: 4a² – b² + 2a + b
Sol: 4a² – b² + 2a + b
= (2a)² – (b)² + 2a + b
= (2a + b) (2a – b) + 1 (2a + b)
= (2a + b) (2a – b + 1)
Que-18: x² + 3x – y² – 3y
Sol: x² + 3x – y² – 3y
= x² – y² + 3x – 3y
= (x + y) (x – y) + 3 (x – y) {∵ a² – b² = (a + b) (a – b)}
= (x – y) (x + y + 3)
Que-19: a² + b² – 2ab – 4c²
Sol: a² + b² – 2ab – 4c²
= (a – b)² – (2c)²
{∵ a² – b² = (a + 6) (a – 6)}
{∵ a² + b² – 2ab = (a – b)²}
= (a – b + 2c) (a – b – 2c)
Que-20: 9x² – 6xy + y² – z²
Sol: 9x² – 6xy + y² – z²
= [(3x)² – 2 x 3x × y + (y)²] – (z)² {∵ a² – 2ab + b² = (a – b)²]
= (3x – y)² – (z)² {∵ a² – b² – (a + b) (a – b)}
= (3x – y + z) (3x – y – z)
Que-21: x² – 1 – 2a – a²
Sol: x² – 1 – 2a – a²
= x² – (1 + 2a + a²)
= (x)² – (1 + a)² {a² + 2ab + b² = (a + b)²}
= (x + 1 + a) (x – 1 – a)
Que-22: 4a² + b² – c² + 4ab
Sol: 4a² + b² – c² + 4ab
= 4a² + b² + 4ab – c²
= (2a)² + (b)² + 2 x 2a x b – (c)² {a² + b² + 2ab = (a + b)²}
= (2a + b)² – (c)²
= (2a + b + c) (2a + b – c)
Que-23: x³ + 2x² – x – 2
Sol: x³ + 2x² – x – 2
= x² (x + 2) – 1 (x + 2)
= (x + 2) (x² – 1)
= (x + 2) {(x)² – (1)²}
= (x + 2) (x + 1) (x – 1)
Que-24: 1 + 2ab – (a² + b²)
Sol: 1 + 2ab – (a² + b²)
= 1 + 2ab – a² – b²
= 1 – (a² + b² – 2ab)
= (1)² – (a – b)²
= (1 + a – b) (1 – a + b)
Que-25: x² + (1/x²) – 11
Sol: x² + (1/x²) – 11
= x² + (1/x²) – 2 – 9
= (x)² – 2 + {1/(x)²} – (3)²
= (x – (1/x))² – (3)²
= (x – (1/x) + 3) (x – (1/x) – 3)
Que-26: x4 + 3x² + 4
Sol: x4 + 3x² + 4 = x4 + 4x² + 4 – x²
= (x²)² + 2 × x² + (2)² – (x)²
= (x² + 2)² – (x)²
= (x² + 2 – x) (x² + 2 + x)
Que-27: Factorise the following :
(i) 4a² – b² + 2a + b
(ii) 9x² – 4 (y + 2x)²
(iii) 9 (x + y)² – x²
Sol: (i) 4a² – b² + 2a + b
= (2a)² – (b)² + 2a + b
= (2a + b) (2a – b) + 1 (2a + b)
= (2a + b) (2a – b + 1)
(ii) 9x² – 4 (y + 2x)² = 9x² – [2 (y + 2x)]²
= (3x)² – [2 (y + 2x)]²
= [3x + 2 (y + 2y)] [3x – 2 (y + 2x)]
= (3x + 2y + 4x) (3x – 2y – 4x)
= (7x + 2y) (- x – 2y)
= – (7x + 2y) (x + 2y)
(iii) 9 (x + y)² – x²
= [3 (x + y)]² – (x)²
= [3 (x + y) + x] [3 (x + y) – x]
= (3x + 3y + x) (3x + 3y – x)
= (4x + 3y) (2x + 3y)
Que-28: Factorise : x³ – 3x² – x + 3
Sol: x² – 3x² – x + 3
= x² (x – 3) – 1 (x – 3)
= (x – 3) (x² – 1)
= (x – 3) [(x)² – (1)²]
= (x – 3) (x + 1) (x – 1)
= (x + 1) (x – 1) (x – 3)
Que-29: Factorise :
(a² – b²) (c² – d²) – 4abcd
Sol: (a² – b²) (c² – d²) – 4abcd
= a²c² – a²d² – b²c² + b²d² – 4abcd
= a²c² + b²d² – 2abcd – a²d² – b²c² – 2abcd
= (ac – bd)² – (ad + bc)²
= (ac – bd+ ad + bc) (ac – bd – ad – bc)
= [ac + ad – bc – bd] [ac – ad – bc – bd]
= [a (c + d) – b (c – d)][a (c – d) – d (c + d)]
Que-30: Express (x² + 8x + 15) (x² – 8x + 15) as a difference of two squares.
Sol: (x² + 8x + 15) (a² – 8x + 15)
= [(x² + 15) + 8x] [(x² + 15) – 8x]
= (x² + 15)² – (8x)²
— : End of Factorisation Class 9 OP Malhotra Exe-4E ICSE Maths Solutions Ch-4 :–
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