The availability of OP Malhotra Class 9 Solutions Chapter 4 Factorisation Ex 4(B) encourages students to tackle difficult exercises.
S Chand Class 9 ICSE Maths Solutions Chapter 4 Factorisation Ex 4(B)
Factorise the following :
Question 1.
bx + 2b + cx + 2c
Solution:
bx + 2b + cx + 2c
= bx + cx + 2b + 2c
= x(b + c) + 2 (b + c)
= (b + c) (x + 2)
Question 2.
y² + 2y² + 3y + 6
Solution:
y² + 2y² + 3y + 6
= y²(y + 2) + 3(y + 2)
= (y + 2) (y² + 3)
Question 3.
xa + 3b + xb + 3a
Solution:
xa + 3b + xb + 3a
= xz + xb + 3a + 3b
= x (a + b) + 3 (a + b)
= (a + b) (x + 3)
Question 4.
8xy + 5zy – 8xt – 5zt
Solution:
8xy + 5zy – 8xt – 5zt
= 8xy – 8xt + 5zy – 5zt
= 8x (y – t) + 5z (y – t)
= (y – r) (8x + 5z)
Question 5.
8kl + 12ml – 12mn – 8kn
Solution:
8kl + 12ml – 2mn – 8kn
= 8kl – 8kn + 12ml – 12mn
= 8k (l – n) + 12m (l – n)
= (l – n) (8k + 12m)
Question 6.
32 (x + y)² – 2x – 2y
Solution:
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)
Question 7.
x³ – x² + ax + x – a – 1
Solution:
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)
Question 8.
ab (c² + 1) + c (a² + b²)
Solution:
ab (c² + 1) + c (a² + b²)
= abc² + ab + a²c + b²c
= abc² + a²c + b²c + ab
= ac (be + a) + b (bc + a)
= (bc + a) (ac + b)
Question 9.
a³ – a² + xa + a – x – 1
Solution:
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)
Question 10.
6a³b + 3a²b² – 2a²b – ab²
Solution:
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)
Question 11.
a³ + ab (1 – 2a) – 2b²
Solution:
a³ + ab (1 – 2d) – 2b²
= a³ + ab – 2 a²b – 2 b²
= a (a² + b) – 2b (a² + b)
= (a² + b) (a- 2b)
Question 12.
(p² + 1)q – p² – q²
Solution:
(p² + 1 )q – p² – q²
= p²q + q – p² – q²
= p²q – p² – q² + q
=p² (q – 1) – q(q – 1)
= (q – 1) (p² – q)