Class 8: Maths Chapter 11 solutions. Complete Class 8 Maths Chapter 11 Notes.
Contents
ML Aggarwal Solutions for Class 8 Maths Chapter 11- Factorisation
ML Aggarwal 8th Maths Chapter 11, Class 8 Maths Chapter 11 solutions
Exercise 11.1
Factorise the following (1 to 8) polynomials:
1. (i) 8xy3 + 12x2y2
(ii) 15ax3 – 9ax2
Solution:
(i) 8xy3 + 12x2y2 = 4xy2 (2y + 3x)
(ii) 15ax3 – 9ax2 = 3ax2 (5x – 3)
2. (i) 21 py2 – 56py
(ii) 4x3 – 6x2
Solution:
(i) 21 py2 – 56py = 7py (3y – 8)
(ii) 4x3 – 6x2 = 2x2 (2x – 3)
3. (i) 25abc2 – 15a2b2c
(ii) x2yz + xy2z + xyz2
Solution:
(i) 25abc2 – 15a2b2c = 5abc (5c – 3ab)
(ii) x2yz + xy2z + xyz2 = xyz(x + y + z)
4. (i) 8x3 – 6x2 + 10x
(ii) 14mn + 22m – 62p
Solution:
(i) 8x3 – 6x2 + 10x = 2x (4x2 – 3x + 5)
(ii) 14mn + 22m – 62p = 2 (7mn + 11m – 31p)
5. (i) 18p2q2 – 24pq2 + 30p2q
(ii) 27a3b3 – 18a2b3 + 75a3b2
Solution:
(i) 18p2q2 – 24 pq2 + 30p2q
= 6pq (3pq -4q + 5p)
(ii) 27a3b3 – 18a2b3 + 75a3b2
= 3a2b2 (9ab – 6b + 25a)
6. (i) 15a (2p – 3p) – 106 (2p – 3q)
(ii) 3a (x2 + y2) + 6b (x2 + y2)
Solution:
(i) 15a (2p – 3q) – 10b (2p – 3q)
= (2p – 3q)(15a – 10b)
= (2p – 3q) (5) (3a – 2b)
= 5 (2p- 3q) (3a – 2b)
(ii) 3a (x2 + y2) + 66 (x2 + y2)
= (x2 + y2) (3a + 6b)
= (x2 + y2) (3) (a + 2b)
= 3 (x2 + y2) (a + 2b)
7. (i) 6(x + 2y)3 + 8(x + 2y)2
(ii) 14(a – 3b)3 – 21p(a – 3b)
Solution:
(i) 6(x + 2y)3 + 8(x + 2y)2
(x + 2y)2 [6 (x + 2y) + 8]
= (x + 2y)2 [6x + 12y + 8]
= (x + 2y)2 (2) (3x + 6y + 4)
= 2 (x + 2y)2 (3x + 6y + 4)
(ii) 14(a – 3b)3 – 21 p(a – 3b)
= 7 [2 (a – 3b)3 – 3p(a- 3b)]
= 7 [(a – 3b) {2 (a – 3b)2 – 3p}]
= 7 (a – 3b) [2 (a – 3b)2 – 3p]
8. 10a (2p + q)3 – 15b (2p + q)2 + 35(2p + q)
Solution:
10a (2p + q)3 – 15b (2p + q)2 + 35(2p + q)
= 5 [2a (2p + q)]3 – 3b (2p + q)2 + 7 (2p + q)
= 5(2p + q) [2a (2p + q)2 – 3b(2p + q) + 7]
Exercise 11.2
Factorise the following (1 to 11) polynomials:
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
Exercise 11.3
1. Factorise the following expressions using algebraic identities:
(i) x2 – 12x + 36
(ii) 36p2 – 60pq + 25q2
(iii) 9y2 + 66xy + 121y2
(iv) a4 + 6a2b2 + 9b4
(v) x2 + 1/x2 + 2
(vi) x2 + x + 1/4
Solution:
Using (a + b)2 = a2 + 2ab +b2 and (a – b)2 = a2 – 2ab + b2
(i) y2 – 12x + 36
= (x)2 – 2 × x × 6 + (6)2
= (x – 6)2
(ii) 36p2 – 60pq + 25q2
= (6p)2 – 2 × 6p × 5q + (5q)2
= (6p – 5q)2
(iii) 9x2 + 66xy + 121 y2
= (3x)2 + 2 × 3x × 11y + (11y)2
= (3x + 11 y)2
(iv) a4 + 6a2b2 + 9b4
= (a2)2 + 2 × a2 × 3b2 + (3b2)2
= (a2 + 3b2)2
(v) x2 + 1/x2 + 2
= (x)2 + 2 × x × 1/x + (1/x)2
= (x + 1/x)2
(vi) x2 + x + ¼
= (x)2 + 2 × x × 1/2 + (1/2)2
= (x + 1/2)2
Factorise the following (2 to 13) expressions:
2. (i) 4p2 – 9
(ii) 4x2 – 169y2
Solution:
(i) 4p2 – 9
= (2p)2 – (3)2
= (2p + 3) (2p – 3)
(ii) 4x2 – 169y2
= (2x)2 – (13y)2
= (2x + 13y) (2x – 13y)
3. (i) 9x2y2 – 25
(ii) 16x2 – 1/144
Solution:
(i) 9x2y2 – 25
= (3xy)2 – (5)2
= (3xy + 5) (3xy – 5)
(ii) 16x2 – 1/144
= (4x)2 – (1/12)2
= (4x + 1/12) (4x – 1/12)
4. (i) 20x2 – 45y2
(ii) 9/16 – 25a2b2
Solution:
(i) 20x2 – 45y2
= 5 (4x2 – 9y2)
= 5[(2x)2 – (3y)2]
= 5 (2x + 3y) (2x – 3y)
(ii) 9/16 – 25a2b2
= (¾)2 – (5ab)2
= (¾ + 5ab) (¾ – 5ab)
5. (i) (2a + 3b)2 – 16c2
(ii) 1 – (b – c)2
Solution:
(i) (2a + 3b)2 – 16c2
= (2a + 3b)2 – (4c)2
= (2a + 3b + 4c) (2a + 3b – 4c)
(ii) 1 – (b – c)2
= (1)2 – (b – c)2
= [1 + b – c)] [1 – (b – c)]
= (1 +b – c) (1 – b + c)
6. (i) 9 (x + y)2 – x2
(ii) (2m + 3n)2 – (3m + 2n)2
Solution:
(i) 9 (x + x)2 – x2
= [3 (x + y)]2 – [x]2
= [3 (x + y) + x] [3 (x + y) – x]
= (3x + 3y + x) (3x + 3y – x)
= (4x + 3y) (2x + 3x)
(ii) (2m + 3n)2 – (3m + 2n)2
= (4m2 + 9n2 + 12mn) – (9m2 + 4n2 + 12mn)
= 4m2 + 9n2 + 12mn – 9m2 – 4m2 – 12mn
= 4m2 + 9n2 – 9m2 – 4n2
= – 5m2 + 5n2
= 5 (n2 – m2)
= 5 (m + n) (n – m)
7. (i) 25 (a + b)2 – 16 (a – b)2
(ii) 9 (3x + 2)2 – 4 (2x – 1)2
Solution:
(i) 25 (a + b)2 – 16 (a – b)2
= [5 (a + b)]2 – [4 (a – b)]2
= (5a + 5b)2 – (4a – 4b)2
= [(5a + 5b)2 + (4a – 4b)] [(5a + 5b) – (4a – 4b)]
= (5a + 5b + 4a – 4b) (5a + 5b – 4a + 4b)
= (9a + b) (a + 9b)
(ii) 9 (3x + 2)2 – 4 (2x – 1)2
= [3 (3x + 2)]2 – [2 (2x – 1)]2
= (9x + 6)2 – (4x – 2)2
= [(9x + 6) + (4x – 2)] [(9x + 6) – (4x – 2)]
= (9x + 6 + 4x – 2) (9x + 6 – 4x + 2)
= (13x + 4) (5x + 8)
8. (i) x3 – 25x
(ii) 63p2q2 – 7
Solution:
(i) x3 – 25x
= x (x2 – 25) = x [(x)2 – (5)2]
= x (x + 5) (x – 5)
(ii) 63p2q2 – 7
= 7 (9p2q2 – 1)
= 7 [(3pq)2 – (1)2]
= 7 (3pq + 1) (3pq – 1)
9. (i) 32a2b – 72b3
(ii) 9 (a + b)3 – 25 (a + b)
Solution:
(i) 32 a2b – 72b3
= 8b (4a2 – 9b2) ⇒ 8b [(2a)2 – (3b)2]
= 8b (2a + 3b) (2a – 3b)
(ii) 9 (a + b)3 – 25 (a + b)
= (a + b) [9 (a + b)2 – 25]
= (a + b) [{3 (a + b)}2 – (5)2]
= (a + 6) [(3a + 3b)2 – (5)2]
= (a + b) [(3a + 3b + 5) (3a + 36 – 5)]
= (a + b) (3a + 3b + 5) (3a + 3b – 5)
10. (i) x2 – y2 – 2y – 1
(ii) p2– 4pq + 4q2 – r2
Solution:
(i) x2 – y2 – 2y – 1
= x2 – (y2 + 2y + 1)
= (x)2 – (y + 1)2
= [x + (y + 1)] [x – (y + 1)]
= (x + y + 1) (x – y – 1)
(ii) p2 – 4pq + 4q2 – r2
= (p)2 – 2 × p × 2q + (2q)2 – r2 [∵ (a – b)2 = a2 – 2ab + b2]
= (p – 2q)2 – (r)2
= (p – 2q + r)(p – 2q – r) [∵ a2 – b2 = (a + b)(a – b)]
11. (i) 9x2 – y2 + 4y – 4
(ii) 4a2 – 4b2 + 4a + 1
Solution:
(i) 9x2 – y2 + 4y – 4
= 9x2 – (y2 – 4y + 4)
= 9x2 – (y – 2)2
= (3x)2 (y – 2)2
= [3x + (y – 2)] [3x – (y – 2)]
= (3x + y – 2) (3x – y + 2)
(ii) 4a2 – 4b2 + 4a + 1
= (4a2 + 4a + 1) – 4b2
= (2a + 1)2 – (2b)2
= (2a + 2b + 1) (2a – 2b + 1)
12. (i) 625 – p4
(ii) 5y5 – 405y
Solution:
(i) 625 – p4
= (25)2 – (p2)2
= (25 + p2) (25 – p2)
= (25 + p2) [(5)2 – (p)2]
= (25 +p2) (5 + p) (5 – p)
(ii) 5y5 – 405y
= 5y(y4 – 81)
= 5y [(y2)2 – (9)2]
= 5y (y2 + 9) (y2 – 9)
= 5y (y2 + 9) [(y)2 – (3)2
= 5y (y2 + 9) (y + 3) (y – 3)
13. (i) x4 – y4 + x2 – y2
(ii) 64a2 – 9b2 + 42bc – 49c2
Solution:
(i) x4 – y4 + x2 – y2
= [(x2)2 – (y2)2] + (x2 – y2) [Using, a2 – b2 = (a + b) (a – b)]
= (x2 + y2) (x2 – y2) + 1(x2 – y2)
= (x2 – y2) (x2 + y2 + 1)
= (x + y(x – y)(x2 + y2 + 1)
(ii) 64a2 – 9b2 + 42bc – 49c2
= 64a2 – [9b2 – 42bc + 49c2]
= (8a)2 – [(3b)2 – 2 × 3b × 7c + (7c)2] [∵ a2 + b2 – 2ab = (a – b)2 and a2 – b2 = (a + b)(a – b)]
= (8a)2 – (3b – 7c)2
= (8a + 3b – 7c) (8a – 3b + 7c)
Exercise 11.4
1. (i) x2 + 3x + 2,
(ii) z2 + 10z + 24
Solution:
(i) x2 + 3x + 2
= x2 + 2x + x + 2
= x(x + 2) + 1 (x + 2)
= (x + 2) (x + 1)
(ii) z2 + 10z + 24
= z2 + 6z + 4z + 24
= z(z + 6) + 4 (z + 6)
= (z + 6) (z + 4)
2. (i) y2 – 7y + 12
(ii) m2 – 23m + 42
Solution:
(i) y2 – 7y + 12
= y2 – 3y – 4y + 12 [Since, 12 = -3 × (-4)1 and – 7 = -3 – 4}
= y(y – 3) -4(y – 3)
= (y – 3) (y – 4)
(ii) m2 – 23m + 42
= m2 – 2m – 21m + 42 [Since, 42 = -2 × (-21) and – 23 = -21 – 2]
= m(m – 2) – 21(m – 2)
= (m – 2) (m – 21)
3. (i) y2 – 5y – 24,
(ii) t2 + 23t – 108
Solution:
(i) y2– 5y – 24
= y2 – 8y + 3y – 24
= y(y – 8) + 3(y – 8)
= (y – 8) (y + 3)
(ii) t2 + 23t – 108
= t2 + 27t – 4t – 108
= t (t + 27) – 4 (t + 27)
= (t + 27) (t – 4)
4. (i) 3x2 + 14x + 8,
(ii) 3y2 + 10y + 8
Solution:
(i) 3x2 + 14x + 8
= 3x2 + 12x + 2x + 8
= 3x (x + 4) + 2 (x + 4)
= (x + 4) (3x + 2)
(ii) 3y2 + 10y + 8
= 3y2 + 6y + 4y + 8
= 3y (y + 2) + 4 (y + 2)
= (y + 2) (3y + 4)
5. (i) 14x2 – 23x + 8,
(ii) 12x2 – x – 35
Solution:
(i) 14x2 – 23x + 8
= 14x2 – 16x – 7x + 8
= 2x (7x – 8) – 1 (7x – 8)
= (7x – 8) (2x – 1)
(ii) 12x2 – x – 35
= 12x2 – 21x + 20x – 35
= 3x (4x – 7) + 5 (4x – 7)
= (4x – 7) (3x + 5)
6. (i) 6x2 + 11x – 10
(ii) 5 – 4x – 12x2
Solution:
(i) 6x2 + 11x – 10
= 6x2 + 15x – 4x – 10
= 3x (2x + 5) – 2 (2x + 5)
= (2x + 5) (3x – 2)
(ii) 5 – 4x- 12x2
= 5 – 10x + 6x – 12x2
= 5 (1 – 2x) + 6x (1 – 2x)
= (1 – 2x) (5 + 6x)
7. (i) 1 – 18y – 63y2,
(ii) 3x2 – 5xy – 12y2
Solution:
(i) 1 – 18y – 63y2
= 1 – 21y + 3y – 63y2
= 1(1- 21y) + 3y (1 – 21y)
= (1 – 21y) (1 + 3y)
(ii) 3x2 – 5xy – 12y2
= 3x2 – 9xy + 4xy- 12y2
= 3x (x – 3y) + 4y (x – 3y)
= (x – 3y) (3x + 4y)
8. (i) x2 – 3xy – 40y2
(ii) 10p2q2 – 21pq + 9
Solution:
(i) x2 – 3xy – 40y2
= x2 – 8xy + 5xy – 40y2
= x (x – 8y) + 5y (x – 8y)
= (x – 8y) (x + 5y)
(ii) 10p2q2 – 21pq + 9
= 10p2q2 – 15pq – 6pq + 9
= 5pq (2pq – 3) – 3 (2pq – 3)
= (2pq – 3) (5pq – 3)
9. (i) 2a2b2 + ab – 45
(ii) x (12x + 7) – 10
Solution:
(i) 2a2b2 + ab – 45
= 2a2b2 + 10ab – 9ab – 45
= 2ab (ab + 5) – 9 (ab + 5)
= (ab + 5) (2ab – 9)
(ii) x (12x + 7) – 10
= 12x2 + 7x – 10
= 12x2 + 15x – 8x – 10
= 3x (4x + 5) – 2 (4x + 5)
= (4x + 5) (3x – 2)
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 we have
= x2 – 11x – 42
= x2 – 14x + 3x – 42 [∵ -42 = -14 × 3 and -11 = -14 + 3]
= x(x – 14) + 3 (x – 14)
= (x – 14) (x + 3)
Substituting the value of x we get,
= (a + b – 14) (a + b + 3)
(ii) 8 + 6(p + q) – 5(p + q)2
Let p + q = x, then we have
= 8 + 6x – 5x2
= -5x2 + 6x + 8
= -(5x2 – 6x – 8)
= 5x2 – 10x + 4x – 8 [∵ 5 × (-8) = 40 ⇒ -40 = -10 × 4 and -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)
11. (i) (x – 2y)2 – 6(x – 2y) + 5
(ii) 7 + 10(2x – 3y) – 8(2x – 3y)2
Solution:
(i) Let x – 2y = z
Then, (x – 2y)2 – 6 (x – 2y) + 5 becomes
= z2 – 6z + 5
= z2 – 5z – z + 5
= z(z – 5) – 1 (z – 5)
= (z – 5)(z – 1)
Now, on 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 becomes
= 7 + 10z – 8z2
= 7 + 14z – 4z – 8z2
= 7 (1 + 2z) – 4z (1 + 2z)
= (1 + 2z) (7 – 4z)
Now, on substituting z = 2x – 3y, we get
= [(1 + 2 (2x – 3y)] [7 – 4 (2x – 3y)]
= (1 + 4x – 6y) (7 – 8x + 12y)
Exercise 11.5
Work out the following divisions:
(i) (35x + 28) ÷ (5x + 4)
(ii) 7p2q2(9r – 27) ÷ 63pq(r – 3)
Solution:
(i) (35x + 28) ÷ (5x + 4)
(ii) 7p2q2(9r – 27) ÷ 63pq(r – 3)
=
= p2-1 q2-1 × 9 = 9pq
2. Divide as directed:
(i) 6(2x + 7) (5x – 3) ÷ 3(5x – 3)
(ii) 33pq (p + 3) (2q – 5) ÷ 11p (2q – 5)
Solution:
(i) 6(2x + 7) (5x – 3) ÷ 3(5x – 3)
=
= 2(2x + 7)
(ii) 33pq (p + 3) (2q – 5) ÷ 11p (2q – 5)
=
= 3q(p + 3)
3. Factorise the expression and divide them as directed:
(i) (7x2 – 63x) ÷ 7(x – 3)
(ii) (3p2 + 17p + 10) ÷ (p + 5)
(iii) 10xy(14y2 + 43y – 21) ÷ 5x(7y – 3)
(iv) 12pqr(6p2 – 13pq + 6q2) ÷ 6pq(2p – 3q)
Solution:
(i) (7x2 – 63x) ÷ 7(x – 3)
(ii) (3p2 + 17p + 10) ÷ (p + 5)
= (3p2 + 17p + 10)/ (p + 5)
(iii) 10xy(14y2 + 43y – 21) ÷ 5x(7y – 3)
(iv) 12pqr(6p2 – 13pq + 6q2) ÷ 6pq(2p – 3q)
Check Your Progress
1. Find the HCF of the given polynomials:
(i) 14pq, 28p2q2
(ii) 8abc, 24ab2, 12a2b
Solution:
(i) 14pq, 28p2q2
HCF of 14, 28 = 14
HCF of 14pq, 28p2q2 = 14pq
(ii) 8abc, 24ab2, 12a2b
HCF of 8, 24, 12 = 4
HCF of 8abc, 24ab2, 12a2b = 4ab
2. Factorise the following:
(i) 10x2 – 18x3 + 14x4
(ii) 5x2y + 10xyz + 15xy2
(iii) p2x2 + c2x2 – ac2 – ap2
(iv) 15(x + y)2 – 5x – 5y
(v) (ax + by)2 + (ay – bx)2
(vi) ax + by + cx + bx + cy + ay
(vii) 49x2 – 70xy + 25y2
(viii) 4a2 + 12ab + 9b2
(ix) 49p2 – 36q2
(x) 100x3 – 25xy2
(xi) x2 – 2xy + y2 – z2
(xii) x8 – y8
(xiii) 12x3 – 14x2 – 10x
(xiv) p2 – 10p + 21
(xv) 2x2 – x – 6
(xvi) 6x2 – 5xy – 6y2
(xvii) x2 + 2xy – 99y2
Solution:
(i) 10x2 – 18x3 + 14x4
HCF of 10, 18, 14 = 2
So, 10x2 – 18x3 + 14x4
= 2x2 (5 – 9x + 7x2)
(ii) 5x2y + 10xyz + 15xy2
HCF of 5, 10, 15 = 5
So, 5x2y + 10xyz + 15xy2
= 5xy (x + 2z + 3y)
(iii) p2x2 + c2x2 – ac2 – ap
= p2x2 – ap2 + c2x2 – ac2
= p2(x2 – a) + c2(x2 – a)
= (x2 – a) (p2 + c2)
(iv) 15(x + y)2 – 5x – 5y
= 15(x + y)2 – 5(x + y)
= 5(x + y) [3(x + y) – 1]
= 5(x + y) (3x + 3y – 1)
(v) (ax + by)2 + (ay – bx)2
On expanding, we have
= a2x2 + b2y2 + 2abxy + a2y2 + b2x2 – 2abxy
= a2x2 + a2y2 + b2x2 + b2y2
= a2(x2 + y2) + b2(x2 + y2)
= (x2 + y2) (a2 + b2)
(vi) ax + by + cx + bx + cy + ay
= ax + bx + cx + ay + by + cy [On grouping the like variables]
= x(a + b + c) + y(a + b + c)
= (a + b + c) (x + y)
(vii) 49x2 – 70xy + 25y2
= (7x)2 – 2 × 7x × 5y + (5y)2 [∵ (a – b)2 = a2 – 2ab + b2]
= (7x – 5y)2
(viii) 4a2 + 12ab + 9b2
= (2a)22 + 2 × 2a × 3b + (3b)2 [∵ (a + b)2 = a2 + 2ab + b2]
= (2a + 3b)2
(ix) 49p2 – 36q2
= (7p)2 – (6q)2
= (7p + 6q) (7p – 6q) [∵ a2 – b2 = (a + b) (a – b)]
(x) 100x3 – 25xy2
= 25x(x2 – y2) = 25x{(x)2 – (y)2}
= 25x(x + y) (x – y)
(xi) x2 – 2xy + y2 – z2
= (x – y)2 – (z)2 [∵ a2 -2ab + b2 = (a – b)2 and a2 – b2 = (a + b) (a – b)]
= (x – y + z)(x – y – z)
(xii) x8 – y8
= (x4)2 – (y4)2 [∵ a2 – b2 = (a + b)(a- b)]
= (x4 + y4) (x4 – y4)
= (x4 + y4) [(x2)2 – (y2)2]
= (x4 + y4) (x2 + y2) (x2 – y2)
= (x4 + y4 (x2 + y2) (x + y) (x – y)
(xiii) 12x3 – 14x2 – 10x
= 2x(6x2 – 7x – 5) [Now, as 6 × (-5) = -30 ⇒ -30 = -10 × 3 and -7 = -10 + 3]
= 2x(6x2 + 3x – 10x – 5)
= 2x{3x(2x + 1) – 5(2x + 1)}
= 2x(2x + 1) (3x – 5)
(xiv) p2 – 10p + 21
= p2 – 3p – 7p + 21 [Now, as 21 =-3 × (-7) and -10 = -3 – 7]
= p(p – 3) – 7(p – 3)
= (p – 3)(p – 7)
(xv) 2x2 – x – 6
= 2x2 – 4x + 3x – 6 [Now, as -6 × 2 = -12 ⇒ -12 = -4 × 3 and -1 = -4 + 3]
= 2x (x – 2) + 3(x – 2)
= (x – 2) (2x + 3)
(xvi) 6x2 – 5xy – 6y2
= 6x2 – 9xy + 4xy – 6y2 [Now, as 6 × (-6) = -36 ⇒ – 36 = -9 × 4 and – 5 = -9 + 4]
= 3x(2x – 3y) + 2y(2x – 3y)
= (2x – 3y) (3x + 2y)
(xvii) x2 + 2xy – 99y2
= x2 + 11xy – 9xy – 99y2 [Now, as -99 = -11 × 9 and -2 = -11 + 9 }
= x(x + 11y) – 9y(x + 11y)
= (x + 11y) (x – 9y)
3. Divide as directed:
(i) 15(y + 3)(y2 – 16) ÷ 5(y2 – y – 12)
(ii) (3x3 – 6x2 – 24x) ÷ (x – 4) (x + 2)
(iii) (x4 – 81) ÷ (x3 + 3x2 + 9x + 27)
Solution:
(i) 15(y + 3)(y2 – 16) ÷ 5(y2 – y – 12)
y2 – 16 = (y)2 – (4)2
= (y + 4)(y – 4)
y2 – y – 12 = y2 – 4y + 3y – 12
= y(y – 4) + 3(y – 4)
= (y – 4)(y + 3)
Now,
= 3(y + 4)
(ii) (3x3 – 6x2 – 24x) ÷ (x – 4) (x + 2)
3x3 – 6x2 – 24x = 3x(x2 – 2x – 8)
= 3x{x2 – 4x + 2x – 8}
= 3x{x(x – 4) + 2(x – 4)}
= 3x(x – 4) (x + 2)
Now,
(iii) (x4 – 81) ÷ (x3 + 3x2 + 9x + 27)
x4 – 81 = (x2)2 – (9)2 = (x2 + 9) (x2 – 9)
= (x2 + 9) [(x)2 – (3)2]
= (x2 + 9) (x + 3) (x – 3)
And,
x3 + 3x2 + 9x + 27 = (x)2 + (x + 3) + 9 (x + 3)
= (x2 + 9) (x + 3)
Now,
= (x – 3)
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ML Aggarwal Solutions for Class 8 Maths Chapter 11- Factorisation
Download PDF: ML Aggarwal Solutions for Class 8 Maths Chapter 11- Factorisation PDF
Chapterwise ML Aggarwal Solutions for Class 8 Maths :
- Chapter 1- Rational Numbers
- Chapter 2- Exponents and Powers
- Chapter 3- Squares and Square Roots
- Chapter 4- Cubes and Cube Roots
- Chapter 5- Playing with Numbers
- Chapter 6- Operation On Sets Venn Diagram
- Chapter 7- Percentage
- Chapter 8- Simple and Compound Interest
- Chapter 9- Direct and Inverse Variation
- Chapter 10- Algebraic Expressions and Identities
- Chapter 11- Factorisation
- Chapter 12- Linear Equations and Inequalities in One Variable
- Chapter 13- Understanding Quadrilaterals
- Chapter 14- Constructions of Quadrilaterals
- Chapter 15- Circle
- Chapter 16- Symmetry Reflection and Rotation
- Chapter 17- Visualising Solid Shapes
- Chapter 18- Mensuration
- Chapter 19- Data Handling
About ML Aggarwal
M. L. Aggarwal, is an Indian mechanical engineer, educator. His achievements include research in solutions of industrial problems related to fatigue design. Recipient Best Paper award, Manipal Institute of Technology, 2004. Member of TSTE.