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

**Factorise the following (1 to 11) polynomials:**

Question 1.

(i) x^{2} + 3x + 2,

(ii) z^{2} + 10z + 24

Solution:

(i) x^{2} + 3x + 2

= x^{2} + 2x + x + 2

= x(x + 2) + 1 (x + 2)

= (x + 2) (x + 1)

(ii) z^{2} + 10z + 24

= z^{2} + 6z + 4z + 24

= z(z + 6) + 4 (z + 6)

= (z + 6) (z + 4)

Question 2.

(i) y^{2} – 7y + 12

(ii) m^{2} – 23m + 42

Solution:

(i) y^{2} – 7y + 12

= y^{2} – 3y – 4y + 12

{∵ 12 = -3 × (-4)1

– 7 = -3 – 4}

= y(y – 3) -4(y – 3)

= (y – 3) (y – 4)

(ii) m^{2} – 23m + 42

= m^{2} – 2m – 21m + 42

{∵ 42 = -2 × (-21)

– 23 = -2 – 2}

= m(m – 2) – 21(m – 2)

= (m – 2) (m – 21)

Question 3.

(i) y^{2} – 5y – 24,

(ii) t^{2} + 23t – 108

Solution:

(i) y^{2}– 5y – 24

= y^{2} – 8y + 3y – 24

= y(y – 8) + 3(y – 8)

= (y – 8) (y + 3)

(ii) t^{2} + 23t- 108

= t^{2} + 27t – 4t – 108

= t (t + 27) – 4 (t + 27)

= (t + 27) (t – 4)

Question 4.

(i) 3x^{2} + 14x + 8,

(ii) 3y^{2} + 10y + 8

Solution:

(i) 3x^{2} + 14x + 8

= 3x^{2} + 12x + 2x + 8

= 3x (x + 4) + 2 (x + 4)

= (x + 4) (3x + 2)

(ii) 3y^{2} + 10y + 8

= 3y^{2} + 6y + 4y + 8

= 3y (y + 2) + 4 (y + 2)

= (y + 2) (3y + 4)

Question 5.

(i) 14x^{2} – 23x + 8,

(ii) 12x^{2} – x – 35

Solution:

(i) 14x^{2} – 23x + 8

= 14x^{2} – 16x – 7x + 8

= 2x (7x – 8) – 1 (7x – 8)

= (7x – 8) (2x – 1)

(ii) 12x^{2} – x – 35

= 12x^{2} – 21x + 20x – 35

= 3x (4x – 7) + 5 (4x – 7)

= (4x – 7) (3x + 5)

Question 6.

(i) 6x^{2} + 11x – 10

(ii) 5 – 4x – 12x^{2}

Solution:

(i) 6x^{2} + 11x – 10

= 6x^{2} + 15x – 4x – 10

= 3x (2x + 5) – 2 (2x + 5)

= (2x + 5) (3x – 2)

(ii) 5 – 4x- 12x^{2}

= 5 – 10x + 6x – 12x^{2}

= 5 (1 – 2x) + 6x(1 – 2x)

= (1 – 2x) (5 + 6x)

Question 7.

(i) 1 – 18y – 63y^{2},

(ii) 3x^{2} – 5xy – 12y^{2}

Solution:

(i) 1 – 18y – 63y^{2}

= 1 – 21y + 3y – 63y^{2}

= 1(1- 21y) + 3y (1 – 21y)

= (1 – 21y) (1 + 3y)

(ii) 3x^{2} – 5xy – 12y^{2}

= 3x^{2} – 9xy + 4xy- 12y^{2}

= 3x (x – 3y) + 4y (x – 3y)

= (x – 3y) (3x + 4y)

Question 8.

(i) x^{2} – 3xy – 40y^{2}

(ii) 10p^{2}q^{2} – 21pq + 9

Solution:

(i) x^{2} – 3xy – 40y^{2}

= x^{2} – 8xy + 5xy – 40y^{2}

= x (x – 8y) + 5y (x – 8y)

= (x – 8y) (x + 5y)

(ii) 10p^{2}q^{2} – 21pq + 9

= 10p^{2}q^{2} – 15pq – 6pq + 9

= 5pq (2pq – 3) – 3 (2pq – 3)

= (2pq – 3) (5pq – 3)

Question 9.

(i) 2a^{2}b^{2} + ab – 45

(ii) x (12x + 7) – 10

Solution:

(i) 2a^{2}b^{2} + ab – 45

2a^{2}b^{2} + 10ab – 9ab – 45

= 2ab (ab + 5) – 9 (ab + 5)

= (ab + 5) (2ab – 9)

(ii) x (12x + 7) – 10

= 12x^{2} + 7x – 10

= 12x^{2} + 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

x^{2} – 11x – 42

x^{2} – 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 – 5x^{2} = -5x^{2} + 6x + 8

= -(5x^{2} – 6x – 8)

= [5x^{2} – 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

= z^{2} – 6z + 5

∴ z^{2} – 6z + 5 = z^{2} – 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 – 8z^{2}

∴ 7 + 10z – 8z^{2} = 7 + 14z – 4z – 8z^{2}

= 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)