## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.4

Factorise the following (1 to 11) polynomials:
Question 1.
(i) x2 + 3x + 2,
(ii) z2 + 10z + 24
Solution:
(i) x2 + 3x + 2
= x2 + 2x + x + 2
= x(x + 2) + 1 (x + 2)
= (x + 2) (x + 1)

(ii) z2 + 10z + 24
= z2 + 6z + 4z + 24
= z(z + 6) + 4 (z + 6)
= (z + 6) (z + 4)

Question 2.
(i) y2 – 7y + 12
(ii) m2 – 23m + 42
Solution:
(i) y2 – 7y + 12
= y2 – 3y – 4y + 12
{∵ 12 = -3 × (-4)1
– 7 = -3 – 4}
= y(y – 3) -4(y – 3)
= (y – 3) (y – 4)

(ii) m2 – 23m + 42
= m2 – 2m – 21m + 42
{∵ 42 = -2 × (-21)
– 23 = -2 – 2}
= m(m – 2) – 21(m – 2)
= (m – 2) (m – 21)

Question 3.
(i) y2 – 5y – 24,
(ii) t2 + 23t – 108
Solution:
(i) y2– 5y – 24
= y2 – 8y + 3y – 24
= y(y – 8) + 3(y – 8)
= (y – 8) (y + 3)

(ii) t2 + 23t- 108
= t2 + 27t – 4t – 108
= t (t + 27) – 4 (t + 27)
= (t + 27) (t – 4)

Question 4.
(i) 3x2 + 14x + 8,
(ii) 3y2 + 10y + 8
Solution:
(i) 3x2 + 14x + 8
= 3x2 + 12x + 2x + 8
= 3x (x + 4) + 2 (x + 4)
= (x + 4) (3x + 2)

(ii) 3y2 + 10y + 8
= 3y2 + 6y + 4y + 8
= 3y (y + 2) + 4 (y + 2)
= (y + 2) (3y + 4)

Question 5.
(i) 14x2 – 23x + 8,
(ii) 12x2 – x – 35
Solution:
(i) 14x2 – 23x + 8
= 14x2 – 16x – 7x + 8
= 2x (7x – 8) – 1 (7x – 8)
= (7x – 8) (2x – 1)

(ii) 12x2 – x – 35
= 12x2 – 21x + 20x – 35
= 3x (4x – 7) + 5 (4x – 7)
= (4x – 7) (3x + 5)

Question 6.
(i) 6x2 + 11x – 10
(ii) 5 – 4x – 12x2
Solution:
(i) 6x2 + 11x – 10
= 6x2 + 15x – 4x – 10
= 3x (2x + 5) – 2 (2x + 5)
= (2x + 5) (3x – 2)

(ii) 5 – 4x- 12x2
= 5 – 10x + 6x – 12x2
= 5 (1 – 2x) + 6x(1 – 2x)
= (1 – 2x) (5 + 6x)

Question 7.
(i) 1 – 18y – 63y2,
(ii) 3x2 – 5xy – 12y2
Solution:
(i) 1 – 18y – 63y2
= 1 – 21y + 3y – 63y2
= 1(1- 21y) + 3y (1 – 21y)
= (1 – 21y) (1 + 3y)

(ii) 3x2 – 5xy – 12y2
= 3x2 – 9xy + 4xy- 12y2
= 3x (x – 3y) + 4y (x – 3y)
= (x – 3y) (3x + 4y)

Question 8.
(i) x2 – 3xy – 40y2
(ii) 10p2q2 – 21pq + 9
Solution:
(i) x2 – 3xy – 40y2
= x2 – 8xy + 5xy – 40y2
= x (x – 8y) + 5y (x – 8y)
= (x – 8y) (x + 5y)

(ii) 10p2q2 – 21pq + 9
= 10p2q2 – 15pq – 6pq + 9
= 5pq (2pq – 3) – 3 (2pq – 3)
= (2pq – 3) (5pq – 3)

Question 9.
(i) 2a2b2 + ab – 45
(ii) x (12x + 7) – 10
Solution:
(i) 2a2b2 + ab – 45
2a2b2 + 10ab – 9ab – 45
= 2ab (ab + 5) – 9 (ab + 5)
= (ab + 5) (2ab – 9)

(ii) x (12x + 7) – 10
= 12x2 + 7x – 10
= 12x2 + 15x – 8x – 10
= 3x (4x + 5) – 2 (4x + 5)
= (4x + 5) (3x – 2)

Question 10.
(i) (a + b)2 – 11(a + b) – 42
(ii) 8 + 6(p + q) – 5(p + q)
Solution:
(i) (a + b)2 – 11(a + b) – 42
Let (a + b) = x, then
x2 – 11x – 42
x2 – 14x + 3x – 42 {∵-42 = -14 × 3
-11 = -14 + 3}
x(x – 14)+ 3 (x – 14)
(x – 14) (x + 3)
Substituting the value of
x = (a + b – 14) (a + b + 3)

(ii) 8 + 6(p + q) – 5(p + q)2
Let p + q = x, then
8 + 6x – 5x2 = -5x2 + 6x + 8
= -(5x2 – 6x – 8)
= [5x2 – 10x + 4x – 8]
{∵ 5 × (-8) = 40
∴ -40 = -10 × 4
-6 = -10 + 4}
= (x – 2) (5x + 4)
Substituting the value of x, then
= -(p + q – 2) (5p + 5q +4)
= (4 + 5p + 5q) (-p – q + 2)
= (4 + 5p + 5 q) (2 – p – q)

Question 11.
(i) (x – 2y)2 – 6(x – 2y) + 5
(ii) 7 + 10(2x – 3y) – 8(2x – 3y)2
Solution:
(i) (x – 2y)2 – 6 (x – 2y) + 5
Let x – 2y = z
Then, (x – 2y)2 – 6 (x – 2y) + 5
= z2 – 6z + 5
∴ z2 – 6z + 5 = z2 – 5z – z + 5
= z(z – 5) – 1 (z – 5)
= (z – 5)(z – 1)
Substituting z = x – 2y, we get,
= [(x-2y) – 5] [(x – 2y) – 1]
= (x – 2y – 5) (x – 2y – 1)

(ii) 7 + 10 (2x – 3y) – 8 (2x – 3y)2
Let 2x – 3y = z
Then, 7 + 10 (2x – 3y) – 8 (2x – 3y)2
= 7 + 10z – 8z2
∴ 7 + 10z – 8z2 = 7 + 14z – 4z – 8z2
= 7 (1 + 2z) – 4z (1 + 2z)
= (1 + 2z) (7 – 4z)
Substituting z = 2x – 3y, we get,
= [(1 + 2 (2x – 3y)] [7 – 4 (2x – 3y)]
= (1 + 4x – 6y) (7 – 8x + 12y)