OP Malhotra Class 9 Maths Solutions Chapter 4 Factorisation Ex 4(E)

Well-structured OP Malhotra Class 9 Solutions Chapter 4 Factorisation Ex 4(E) facilitate a deeper understanding of mathematical principles.

S Chand Class 9 ICSE Maths Solutions Chapter 4 Factorisation Ex 4(E)

Question 1.
7x² – 7
Solution:
7x² – 7 = 7 (x² – 1)
= 7[(x)² – (1)²] = 7(x + 1)(x – 1)
{∵ a² – b² = (a + b) (a – b)}

Question 2.
8 – 50y²z²
Solution:
8 – 50y²z² = 2 (4 – 25y²z²)
= 2[(2)² – (5yz)²]
= 2 (2 + 5yz) (2 – 5yz) {∵ a² – b² = (a + b) (a – b)}

Question 3.
ab² – ac²
Solution:
ab² – ac² = a (b² – c²) = a [(b)² – (c)²]
= a (b + c) (b – c) {∵ a² – b² = (a + b) (a – b)}

Question 4.
36x³ – x
Solution:
36x³ – x = x (36x² – 1)
= x[(6x)²-(l)²]
= x (6x + 1) (6x – 1) (∵ a² – b² = (a + b) (a – b)}

Question 5.
x (x² – 1) + 7 (x² – 1)
Solution:
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)}

Question 6.
t² (t – 3)² – (t – 3)²
Solution:
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)}

Question 7.
5c² (c + 2)² – 45 (c + 2)²
Solution:
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)}

Question 8.
(a + b)² – 1
Solution:
(a + b)² – 1 = (a + b)² – (1)²
= (a + b + 1) (a + b – 1) {∵ a² – b² = (a + b) (a – b)}

Question 9.
1 – (x – y)²
Solution:
1 – (x – y)² = (1)² – (x – y)²
= (1 + x – y) (1 – x + y) {∵ a² – b² = (a + b) (a – b)}

Question 10.
25a² – 16 (x – y)²
Solution:
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)

Question 11.
20 – 45 (m + n)²
Solution:
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]

Question 12.
x⁴ – y⁴
Solution:
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)

Question 13.
x⁴ – 625
Solution:
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)

Question 14.
xy⁵ – yx⁵
Solution:
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)

Question 15.
81x⁴ – 256y⁴
Solution:
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)

Question 16.
a² + ac + bc – b²
Solution:
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)

Question 17.
4a² – b² + 2a + b
Solution:
4a² – b² + 2a + b
= (2a)² – (b)² + 2a + b
= (2a + b) (2a – b) + 1 (2a + b)
= (2a + b) (2a – b + 1)

Question 18.
x² + 3x – y² – 3y
Solution:
x² + 3x – y² – 3y
= x² – y² + 3x – 3y = (x + y) (x – y) + 3 (x – y) {∵ a² – b² = (a + 6) (a – 6)}
= (x – y) (x + y + 3)

Question 19.
a² + b² – 2ab – 4c²
Solution:
a² + b² – 2ab – 4c²
= (a – b)² – (2c)²

= (a – b + 2c) (a – b – 2c)

Question 20.
9x² – 6xy + y² – z²
Solution:
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)

Question 21.
x² – 1 – 2a – a²
Solution:
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)

Question 22.
4a² + b² – c² + 4ab
Solution:
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)

Question 23.
x³ + 2x² – x – 2
Solution:
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)

Question 24.
1 + 2ab – (a² + b²)
Solution:
1 + 2ab – (a² + b²)
= 1 + 2ab – a² – b²
= 1 – (a² + b² – 2ab)
= (1)² – (a – b)²
= (1 + a – b) (1 – a + b)

Question 25.
x² + \(\frac { 1 }{ x² }\) – 11
Solution:
x² + \(\frac { 1 }{ x² }\) – 11
= x² + \(\frac { 1 }{ x² }\) – 2 – 9
= (x)² – 2 + \(\frac { 1 }{ (x)² }\) – (3)²
= (x – \(\frac { 1 }{ x }\))² – (3)²
= (x – \(\frac { 1 }{ c }\) + 3) (x – \(\frac { 1 }{ x }\) – 3)

Question 26.
x⁴ + 3x² + 4
Solution:
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)

Question 27.
Factorise the following :
(i) 4a² – b² + 2a + b
(ii) 9x² – 4 (y + 2x)²
(iii) 9 (x + y)² – x²
Solution:
(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)

Question 28.
Factorise : x³ – 3x² – x + 3
Solution:
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)

Question 29.
Factorise :
(a² – b²) (c² – d²) – 4abcd
Solution:
(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)]

Question 30.
Express (x² + 8x + 15) (x² – 8x + 15) as a difference of two squares.
Solution:
(x² + 8x + 15) (a² – 8x + 15)
= [(x² + 15) + 8x] [(x² + 15) – 8x]
= (x² + 15)² – (8x)²