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Chapter 4

Factorisation — Exercise 4.2

Class - 9 ML Aggarwal Understanding ICSE Mathematics



Exercise 4.2

Question 1(i)

Factorise the following:

x2 + xy - x - y

Answer

x2 + xy - x - y

= x(x + y) - 1(x + y)

= (x + y)(x - 1).

Hence, x2 + xy - x - y = (x - 1)(x + y).

Question 1(ii)

Factorise the following:

y2 - yz - 5y + 5z

Answer

y2 - yz - 5y + 5z

= y(y - z) - 5(y - z)

= (y - z)(y - 5).

Hence, y2 - yz - 5y + 5z = (y - z)(y - 5)

Question 2(i)

Factorise the following:

5xy + 7y - 5y2 - 7x

Answer

5xy + 7y - 5y2 - 7x

Rearranging the terms:

= 5xy - 7x + 7y - 5y2

= x(5y - 7) - y(5y - 7)

= (x - y)(5y - 7)

Hence, 5xy + 7y - 5y2 - 7x = (x - y)(5y - 7).

Question 2(ii)

Factorise the following:

5p2 - 8pq - 10p + 16q

Answer

5p2 - 8pq - 10p + 16q

= 5p2 - 10p - 8pq + 16q

= 5p(p - 2) - 8q(p - 2)

= (5p - 8q)(p - 2).

Hence, 5p2 - 8pq - 10p + 16q = (5p - 8q)(p - 2)

Question 3(i)

Factorise the following:

a2b - ab2 + 3a - 3b

Answer

a2b - ab2 + 3a - 3b

= ab(a - b) + 3(a - b)

= (ab + 3)(a - b).

Hence, a2b - ab2 + 3a - 3b = (ab + 3)(a - b).

Question 3(ii)

Factorise the following:

x3 - 3x2 + x - 3

Answer

x3 - 3x2 + x - 3

= x2(x - 3) + 1(x - 3)

= (x2 + 1)(x - 3).

Hence, x3 - 3x2 + x - 3 = (x2 + 1)(x - 3).

Question 4(i)

Factorise the following:

6xy2 - 3xy - 10y + 5

Answer

6xy2 - 3xy - 10y + 5

= 3xy(2y - 1) - 5(2y - 1)

= (2y - 1)(3xy - 5).

Hence, 6xy2 - 3xy - 10y + 5 = (2y - 1)(3xy - 5).

Question 4(ii)

Factorise the following:

3ax - 6ay - 8by + 4bx

Answer

3ax - 6ay - 8by + 4bx

= 3ax - 6ay + 4bx - 8by

Taking out common terms we get,

3ax - 6ay + 4bx - 8by

= 3a(x - 2y) + 4b(x - 2y)

= (x - 2y)(3a + 4b).

Hence, 3ax - 6ay - 8by + 4bx = (x - 2y)(3a + 4b).

Question 5(i)

5px - 8qy + 4qx - 10py

Answer

Given,

⇒ 5px - 8qy + 4qx - 10py

⇒ 5px + 4qx - 8qy - 10py

⇒ x(5p + 4q) - 2y(4q + 5p)

⇒ (5p + 4q)(x - 2y).

Hence, 5px - 8qy + 4qx - 10py = (5p + 4q)(x - 2y).

Question 5(ii)

9a2y - 9ay + 6a - 6

Answer

Given,

⇒ 9a2y - 9ay + 6a - 6

⇒ 9ay(a - 1) + 6(a - 1)

⇒ (a - 1)(9ay + 6)

⇒ (a - 1).3.(3ay + 2)

⇒ 3(a - 1)(3ay + 2).

Hence, 9a2y - 9ay + 6a - 6 = 3(a - 1)(3ay + 2).

Question 6(i)

Factorise the following:

1 - a - b + ab

Answer

1 - a - b + ab

= 1(1 - a) - b(1 - a)

= (1 - a)(1 - b).

Hence, 1 - a - b + ab = (1 - a)(1 - b).

Question 6(ii)

Factorise the following:

a(a - 2b - c) + 2bc

Answer

a(a - 2b - c) + 2bc

= a2 - 2ab - ac + 2bc

= a2 - ac - 2ab + 2bc

= a(a - c) - 2b(a - c)

= (a - 2b)(a - c).

Hence, a(a - 2b - c) + 2bc = (a - 2b)(a - c).

Question 7(i)

Factorise the following:

x2 + xy(1 + y) + y3

Answer

x2 + xy(1 + y) + y3

= x2 + xy + xy2 + y3

= x(x + y) + y2(x + y)

= (x + y)(x + y2).

Hence, x2 + xy(1 + y) + y3 = (x + y)(x + y2).

Question 7(ii)

Factorise the following:

y2 - xy(1 - x) - x3

Answer

y2 - xy(1 - x) - x3

= y2 - xy + x2y - x3

= y(y - x) + x2(y - x)

= (y - x)(y + x2).

Hence, y2 - xy(1 - x) - x3 = (y - x)(y + x2).

Question 8(i)

Factorise the following:

ab2 + (a - 1)b - 1

Answer

ab2 + (a - 1)b - 1

= ab2 + ab - b - 1

= ab(b + 1) - 1(b + 1)

= (b + 1)(ab - 1)

Hence, ab2 + (a - 1)b - 1 = (b + 1)(ab - 1).

Question 8(ii)

Factorise the following:

2a - 4b - xa + 2bx

Answer

2a - 4b - xa + 2bx

= 2(a - 2b) - x(a - 2b)

= (2 - x)(a - 2b).

Hence, 2a - 4b - xa + 2bx = (2 - x)(a - 2b).

Question 9(i)

Factorise the following:

5ph - 10qk + 2rph - 4qrk

Answer

5ph - 10qk + 2rph - 4qrk

= 5(ph - 2qk) + 2r(ph - 2qk)

= (5 + 2r)(ph - 2qk).

Hence, 5ph - 10qk + 2rph - 4qrk = (5 + 2r)(ph - 2qk).

Question 9(ii)

Factorise the following:

x2 - x(a + 2b) + 2ab

Answer

x2 - x(a + 2b) + 2ab

= x2 - xa - 2bx + 2ab

= x(x - a) - 2b(x - a)

= (x - a)(x - 2b).

Hence, x2 - x(a + 2b) + 2ab = (x - a)(x - 2b).

Question 10(i)

Factorise the following:

ab(x2 + y2) - xy(a2 + b2)

Answer

ab(x2 + y2) - xy(a2 + b2)

= abx2 + aby2 - xya2 - xyb2

= abx2 - xyb2 - xya2 + aby2

= bx(ax - by) - ay(ax - by)

= (bx - ay)(ax - by).

Hence, ab(x2 + y2) - xy(a2 + b2) = (bx - ay)(ax - by).

Question 10(ii)

Factorise the following:

(ax + by)2 + (bx - ay)2.

Answer

(ax + by)2 + (bx - ay)2

= a2x2 + b2y2 + 2axby + b2x2 + a2y2 - 2axby

= a2x2 + b2x2 + b2y2 + a2y2

= x2(a2 + b2) + y2(b2 + a2)

= (x2 + y2)(a2 + b2).

Hence, (ax + by)2 + (bx - ay)2 = (x2 + y2)(a2 + b2).

Question 11(i)

Factorise the following:

a3 + ab(1 - 2a) - 2b2.

Answer

a3 + ab(1 - 2a) - 2b2

= a3 + ab - 2a2b - 2b2

= a3 - 2a2b + ab - 2b2

= a2(a - 2b) + b(a - 2b)

= (a2 + b)(a - 2b).

a3 + ab(1 - 2a) - 2b2 = (a2 + b)(a - 2b).

Question 11(ii)

Factorise the following:

3x2y - 3xy + 12x - 12.

Answer

3x2y - 3xy + 12x - 12.

= 3(x2y - xy + 4x - 4)

= 3[xy(x - 1) + 4(x - 1)]

= 3(x - 1)(xy + 4)

Hence, 3x2y - 3xy + 12x - 12 = 3(x - 1)(xy + 4).

Question 12

Factorise the following:

a2b + ab2 - abc - b2c + axy + bxy.

Answer

a2b + ab2 - abc - b2c + axy + bxy

= ab(a + b) - bc(a + b) + xy(a + b)

= (a + b)(ab - bc + xy).

Hence, a2b + ab2 - abc - b2c + axy + bxy = (a + b)(ab - bc + xy).

Question 13

Factorise the following:

ax2 - bx2 + ay2 - by2 + az2 - bz2.

Answer

ax2 - bx2 + ay2 - by2 + az2 - bz2

= x2(a - b) + y2(a - b) + z2(a - b)

= (a - b)(x2 + y2 + z2).

Hence, ax2 - bx2 + ay2 - by2 + az2 - bz2 = (a - b)(x2 + y2 + z2).

Question 14

Factorise the following:

x - 1 - (x - 1)2 + ax - a.

Answer

x - 1 - (x - 1)2 + ax - a

= (x - 1) - (x - 1)2 + a(x - 1)

= (x - 1)[1 - (x - 1) + a]

= (x - 1)(1 - x + 1 + a)

= (x - 1)(a - x + 2).

Hence, x - 1 - (x - 1)2 + ax - a = (x - 1)(a - x + 2).

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