ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.2
Factorise the following (1 to 11) polynomials:
Question 1.
(i) x2 + xy – x – y
(ii) y2 – yz – 5y + 5z
Solution:
(i) x2 + xy – x – y
= x (x + y) -1 (x + y) = (x + y)(x – 1)
(ii) y2 – yz – 5y + 5z
= y(y – z) -5(y – z)
= (y – z)(y – 5)
Question 2.
(i) 5xy + 7y – 5y2 – 7x
(ii) 5p2 – 8pq – 10p + 16q
Solution:
(i) 5xy + 7y – 5y2 – 7x
= 5xy – 5y2 + 7y – 7x
= 5y(x – y) -7 (x – y)
= (x – y)(5y – 1)
(ii) 5p2 – 8pq – 10p + 16q
= 5p2 – 10p – 8pq + 16q
= 5p (p – 2) – 8q (p – 2)
= (p – 2) (5p – 5q)
= (5p – 8q)(p – 2)
Question 3.
(i) a2b – ab2 + 3a – 3b
(ii) x3 – 3x2 + x – 3
Solution:
(i) a2b – ab2 + 3a – 3b
= ab (a – b) + 3 (a – b) = (a – b) (ab + 3)
(ii) x3 – 3x2 + x – 3
= x2 (x – 3) + 1 (x – 3)
= (x – 3) (x2 + 1)
Question 4.
(i) 6xy2 – 3xy – 10y + 5
(ii) 3ax – 6ay – 8by + 4bx
Solution:
(i) 6xy2 – 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) x2 + xy (1 + y) + y3
(ii) y2 – xy (1 – x) – x3
Solution:
(i) x2 + xy (1 + y) + y3
= x2 + xy + xy2 + y3
= x(x + y) + y2(x + y)
= (x + y) (x + y2)
(ii) y2 – xy (1 – x) – x3
= y2 – xy + x2y – x3
= y(y – x) + x2 (y – x)
= (y – x) (y + x2)
Question 6.
(i) ab2 + (a – 1) b – 1
(ii) 2a – 4b – xa + 2bx
Solution:
(i) ab2 + (a – 1) b – 1
= ab2 + 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) x2 – x(a + 2b) + 2a2
Solution:
(i) 5ph – 10qk + 2rph – 4qrk
= 5 (ph – 2qk) + 2r (ph – 2qk)
= (ph – 2qk) (5 + 2r)
(ii) x2 – x(a + 2b) + 2ab
= x2 – xa – 2bx + 2ab
= x(x – a) – 2b(x – a)
= (x – a) (x – 2b)
Question 8.
(i) ab (x2 + y2) – xy (a2 + b2)
(ii) (ax + by)2 + (bx – ay)2
Solution:
(i) ab (x2 + y2) – xy (a2 + b2)
= abx2 + aby2 – a2xy – b2xy
= (abx2 – b2xy) + (aby2 – a2xy)
= bx (ax – by) – ay (ax – by)
= (ax – by) (bx – ay)
(ii) (ax + by)2 + (bx – ay)2
= (a2x2 + b2y2 + 2abxy) + (b2x2 + a2y2 – 2abxy)
= a2x2 + b2y2 + 2abxy + b2x2 + a2y2 – 2abxy
= a2x2 + b2y2 + b2x2 + a2y2
= a2x2 + a2y2 + b2x2 + a2y
= a2 (x2 + y2) + b2 (x2 + y2)
= (a2 + b2) (x2 + y2)
Question 9.
(i) a3 + ab(1 – 2a) – 2b2
(ii) 3x2y – 3xy + 12x – 12
Solution:
(i) a3 + ab – 2a2b – 2b2
= a3 + ab – 2a2b – 2b2
= a (a2 + b) – 2b (a2 + b)
= (a2 + b) (a – 2b)
(ii) 3x2y – 3xy + 12x- 12
= 3 (x2y – xy + 4x – 4)
= 3 [xy (x – 1) +4 (x – 1)]
= 3 (x – 1) (xy + 4)
Question 10.
(i) a2b + ab2 – abc – b2c + axy + bxy
(ii) ax2 – bx2 + ay2 – by2 + az2 – bz2
Solution:
(i) a2b + ab2 – abc – b2c + axy + bxy
= ab (a + b) – bc (a + b) + xy (a + b)
= (a + b) (ab – bc + xy)
(ii) ax2 – bx2 + ay2 – by2 + az2 – bz2
= x2 (a – b) + y2 (a – b) + z2 (a – b)
= (a – b)(x2 + y2 + z2)
Question 11.
(i) x – 1 – (x – 1)2 + ax – a
(ii) ax + a2x + 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 + a2x + aby + by – (ax + by)2
= (ax + by) + (a2x + aby) – (ax + by)2
= (ax + by) + a (ax + by) – (ax + by)2
= (ax + by) [1 + a – (ax + by)]
= (ax + by) (1 + a – ax – by)