ML Aggarwal Solutions Class 9 Math 4th Chapter Factorisation Exercise 4.2
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Fourth Chapter Factorisation Exercise 4.2. APC Solution Class 9 Exercise 4.2.
(1) (i) x2 + xy – x – y
Solution:
x2 + xy – x – y
= x (x + y) – 1 (x + y)
= (x + y) (x – 1)
(ii) y2– yz – 5y + 5z
Solution:
y2 – yz – 5y + 5z
= y (y – z) – 5 (y – z)
= (y – 5) (y – z)
(2) (i) 5xy + 7y – 5y2 – 7x
Solution:
5xy + 7y – 5y2 – 7x
= 5xy – 5y2 – 7x + 7y
= 5y (x – y) -7 (x – y)
= (5y – 7) (x – y)
(ii) 5p2 – 8pq – 10p + 16q
Solution:
5p2 – 8pq – 10p + 16q
= p (5p – 8q) – 2 (5p – 8q)
= (5p – 8q) (p – 1)
(3) (i) a2b – ab2 + 3a – 3b
Solution:
a2b – ab2 + 3a – 3b
= ab (a – b) + 3 (a – b)
= (ab + 3) (a – b)
(ii) x3 – 3x2 + x – 3
Solution:
x3 – 3x2 + x – 3
= x2 (x – 3) + 1 (x – 3)
= (x2 + 1) (x – 3)
(4) (i) 6xy2 – 3xy – 10y + 5
Solution:
6xy2 – 3xy – 10y + 5
= 3xy (2y – 1) – 5 (2y – 1)
= (3xy – 5) (2y – 1)
(ii) 3ax – 6ay – 8by + 4bx
Solution:
3ax – 6xy – 8by + 4bx
= 3a (x – 2y) + 4b (x – 2y)
= (3a + 4b) (x – 2y)
(5) (i) 1 – a – b + ab
Solution:
1 – a – b + ab
= 1 (1 – a) – b (1 – a)
= (1 – b) (1 – a)
(ii) a (a – 2b – c) + 2bc
Solution:
a2 (a – 2b – c) + 2bc
= a2 – 2ab – ac + 2bc
= a2 – ac – 2ab + 2bc
= a (a – c) – 2b (a – c)
= (a – 2b) (a – c)
(6) (i) x2 + xy (1 + y) + y3
Solution:
x2 + xy (1 + y) + y3
= x2 + xy + xy2 + y3
= x (x + y) + y2 (x + y)
= (x + y)2 (x + y)
(ii) y2 – xy (1 – x) – x3
Solution:
y2 – xy (1 – x) – x3
= y2 – xy + x2y – x3
= y (y – x) + x2 (y – x)
= (y + x2) (y – x)
(7) (i) ab2 + (a – 1) b – 1
Solution:
ab2 + (a – 1) b – 1
= ab2 + ab – b – 1
= ab (b + 1) – 1 (b + 1)
= (ab – 1) (b + 1)
(ii) 2a – 4b – xa + 2bx
Solution:
2a – 4b – xa + 2bx
= 2a – xa – 4b + 2bx
= a (2 – x) -2b (2 – x)
= (a – 2b) (2 – x)
(8) (i) 5ph – 10qk + 2rph – 4qrk
Solution:
5ph – 10qk + 2rph – 4qrk
= 5 (ph – 2qk) + 2k (ph – 2qk)
= (5 + 2r) (ph – 2qk)
(ii) x2 – x (a + 2b) + 2ab
Solution:
x2 – x (a + 2b) + 2ab
= x2 – xa – 2xb + 2ab
= x (x – a) – 2b (x – a)
= (x – 2b) (x – a)
(9) (i) ab (x2 + y2) – xy (a2 + b2)
Solution:
ab (x2 + y2) – xy (a2 + b2)
= abx2 + aby2 – xya2 – xyb2
= abx2 – xya2 – 2yb2 + aby2
= ax (bx – ya) – by (bx – yx)
= (ax – by) (bx – ya)
(ii) (ax + by)2 + (bx – ay)2
Solution:
(ax + by)2 + (bx – ay)2
= a2x2 + 2axby + b2y2 + b3x2 – 2axby + a2y2
= a2x2 + b2x2 + a2y2 + b2y2
= x2 (a2 + b2) +y2 (a2+b2)
= (x2+y2) (a2+b2)
(10) (i) a3 + ab (1 – 2a) – 2b2
Solution:
a3 + ab (1 – 2a) – 2b2
= a3 + ab – 2a2b – 2b2
= a (a2 + b) – 2b (a2 + b)
= (a – 2b) (a2 + b)
(ii) 3x2y – 3xy + 12x – 12
Solution:
3x2y – 3xy + 12x – 12
= 3 (x2y – xy + 4x – 4)
= 3 [xy (x – 1) + 4 (x – 1)]
= 3 (xy + 4) (x – 1)
(11) a2b + ab2 – abc – b2c + axy + bxy
Solution:
a2b + ab2 – abc – b2c + axy + bxy
= ab (a + b) – bc (a + b) + 2y (a + b)
= (ab – bc + xy) (a + b)
(12) ax2 – bx2 + ay2 – by2 + az2 – bz2
Solution:
ax2 – bx2 + ay2 – by2 + az2 – bz2
= x2 (a – b) + y2 (a – b) + z2 (a – b)
= (x2 + y2 + z2) (a – b)
(13) x – 1 – (x – 1)2 + ax – a
Solution:
x – 1 (x – 1)2 + ax – a
= 1 (x – 1) – (x – 1) (x – 1) + a (x – 1)
= (1 – x + 1 + a) (x – 1)
= (2 + a – x) (x – 1)