## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.2

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

(i) x

^{2}+ xy – x – y

(ii) y

^{2}– yz – 5y + 5z

Solution:

(i) x

^{2}+ xy – x – y

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

(ii) y

^{2}– yz – 5y + 5z

= y(y – z) -5(y – z)

= (y – z)(y – 5)

Question 2.

(i) 5xy + 7y – 5y^{2} – 7x

(ii) 5p^{2} – 8pq – 10p + 16q

Solution:

(i) 5xy + 7y – 5y^{2} – 7x

= 5xy – 5y^{2} + 7y – 7x

= 5y(x – y) -7 (x – y)

= (x – y)(5y – 1)

(ii) 5p^{2} – 8pq – 10p + 16q

= 5p^{2} – 10p – 8pq + 16q

= 5p (p – 2) – 8q (p – 2)

= (p – 2) (5p – 5q)

= (5p – 8q)(p – 2)

Question 3.

(i) a^{2}b – ab^{2} + 3a – 3b

(ii) x^{3} – 3x^{2} + x – 3

Solution:

(i) a^{2}b – ab^{2} + 3a – 3b

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

(ii) x^{3} – 3x^{2} + x – 3

= x^{2} (x – 3) + 1 (x – 3)

= (x – 3) (x^{2} + 1)

Question 4.

(i) 6xy^{2} – 3xy – 10y + 5

(ii) 3ax – 6ay – 8by + 4bx

Solution:

(i) 6xy^{2} – 3xy – 10y + 5

3xy(2y – 1) -5(2y – 1)

= (2y – 1) (3xy – 5)

(ii) 3ax – 6ay – 8by + 4bx

= 3ax – 6ay + 4bx – 8by

= 3a (x – 2y) + 4b (x – 2y)

= (x – 2y) (3a + 4b)

Question 5.

(i) x^{2} + xy (1 + y) + y^{3}

(ii) y^{2} – xy (1 – x) – x^{3}

Solution:

(i) x^{2} + xy (1 + y) + y^{3}

= x^{2} + xy + xy^{2} + y^{3}

= x(x + y) + y^{2}(x + y)

= (x + y) (x + y^{2})

(ii) y^{2} – xy (1 – x) – x^{3}

= y^{2} – xy + x^{2}y – x^{3}

= y(y – x) + x^{2} (y – x)

= (y – x) (y + x^{2})

Question 6.

(i) ab^{2} + (a – 1) b – 1

(ii) 2a – 4b – xa + 2bx

Solution:

(i) ab^{2} + (a – 1) b – 1

= ab^{2} + ab – b – 1

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

= (b + 1) (ab – 1)

(ii) 2a – 4b – xa + 2bx

= 2 (a – 2b) -x (a – 2b)

= (a – 2b) (2 – x)

Question 7.

(i) 5ph – 10qk + 2rph – 4qrk

(ii) x^{2} – x(a + 2b) + 2a^{2}

Solution:

(i) 5ph – 10qk + 2rph – 4qrk

= 5 (ph – 2qk) + 2r (ph – 2qk)

= (ph – 2qk) (5 + 2r)

(ii) x^{2} – x(a + 2b) + 2ab

= x^{2} – xa – 2bx + 2ab

= x(x – a) – 2b(x – a)

= (x – a) (x – 2b)

Question 8.

(i) ab (x^{2} + y^{2}) – xy (a^{2} + b^{2})

(ii) (ax + by)^{2} + (bx – ay)^{2}

Solution:

(i) ab (x^{2} + y^{2}) – xy (a^{2} + b^{2})

= abx^{2} + aby^{2} – a^{2}xy – b^{2}xy

= (abx^{2} – b^{2}xy) + (aby^{2} – a^{2}xy)

= bx (ax – by) – ay (ax – by)

= (ax – by) (bx – ay)

(ii) (ax + by)^{2} + (bx – ay)^{2}

= (a^{2}x^{2} + b^{2}y^{2} + 2abxy) + (b^{2}x^{2} + a^{2}y^{2} – 2abxy)

= a^{2}x^{2} + b^{2}y^{2} + 2abxy + b^{2}x^{2} + a^{2}y^{2} – 2abxy

= a^{2}x^{2} + b^{2}y^{2} + b^{2}x^{2} + a^{2}y^{2}

= a^{2}x^{2} + a^{2}y^{2} + b^{2}x^{2} + a^{2}y

= a^{2} (x^{2} + y^{2}) + b^{2} (x^{2} + y^{2})

= (a^{2} + b^{2}) (x^{2} + y^{2})

Question 9.

(i) a^{3} + ab(1 – 2a) – 2b^{2}

(ii) 3x^{2}y – 3xy + 12x – 12

Solution:

(i) a^{3} + ab – 2a^{2}b – 2b^{2}

= a^{3} + ab – 2a^{2}b – 2b^{2}

= a (a^{2} + b) – 2b (a^{2} + b)

= (a^{2} + b) (a – 2b)

(ii) 3x^{2}y – 3xy + 12x- 12

= 3 (x^{2}y – xy + 4x – 4)

= 3 [xy (x – 1) +4 (x – 1)]

= 3 (x – 1) (xy + 4)

Question 10.

(i) a^{2}b + ab^{2} – abc – b^{2}c + axy + bxy

(ii) ax^{2} – bx^{2} + ay^{2} – by^{2} + az^{2} – bz^{2}

Solution:

(i) a^{2}b + ab^{2} – abc – b^{2}c + axy + bxy

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

= (a + b) (ab – bc + xy)

(ii) ax^{2} – bx^{2} + ay^{2} – by^{2} + az^{2} – bz^{2}

= x^{2} (a – b) + y^{2} (a – b) + z^{2} (a – b)

= (a – b)(x^{2} + y^{2} + z^{2})

Question 11.

(i) x – 1 – (x – 1)^{2} + ax – a

(ii) ax + a^{2}x + aby + by – (ax + by)^{2}

Solution:

(i) 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) (2 – x + a)

(ii) ax + a^{2}x + aby + by – (ax + by)^{2}

= (ax + by) + (a^{2}x + aby) – (ax + by)^{2}

= (ax + by) + a (ax + by) – (ax + by)^{2}

= (ax + by) [1 + a – (ax + by)]

= (ax + by) (1 + a – ax – by)