GSEB Solutions for Class 8 Mathematics – Factorization-2 (English Medium)
GSEB SolutionsMathsScience
Exercise
Solution 1:
- 36 – x2 = (6 + x)(6 – x)
36 – x2 = (6)2 – (x)2
= (6 + x)(6 – x) - a2 – b2c2 = (a – bc)(a + bc)
a2 – b2c2 = (a)2 – (bc)2
= (a – bc)(a + bc) - x3 – 49x = x (x + 7)(x – 7)
x3 – 49x = x(x2 – 49)
= x(x2 – 72)
= x(x + 7)(x – 7) - 4x2 – 25 = (2x + 5)(2x – 5)
4x2 – 25 = (2x)2 – (5)2
= (2x + 5)(2x – 5) - x2 + 5x + 6 = (x + 3)(x + 2)
x2 + 5x + 6 = x2 + 2x + 3x + 6
= x(x + 2) + 3(x + 2)
= (x + 3)(x + 2) - x2 – x – 12 = (x + 3)(x – 4)
x2 – x – 12 = x2 – 4x + 3x – 12
= x(x – 4) + 3(x – 4)
= (x + 3)(x – 4) - a3 – 1 = (a – 1)(a2 + a + 1)
a3 – 1 = a3 – 13
= (a – 1)[a2 + (a)(1) + 12]
= (a – 1)(a2 + a + 1) - m3 + 125 = (m + 5)(m2 – 5m + 25)
m3 + 125 = m3 + 53
= (m + 5)[m2 – (m)(5) + 52]
= (m + 5)(m2 – 5m + 25)
Solution 2(1):
16a2b2 – 36
= 4(4a2b2 – 9)
= 4[(2ab)2 – (3)2]
= 4(2ab – 3)(2ab + 3)
Solution 2(2):
625 – 64x2
= (25)2 – (8x)2
= (25 + 8x)(25 – 8x)
Solution 2(3):
4x5 – 64x
= 4x(x4 – 16)
= 4x[(x2)2 – (4)2]
= 4x(x2 + 4)(x2 – 4)
= 4x(x2 + 4)(x + 2)(x – 2)
Solution 2(4):
(4a – 5b)2 – 16c2
= (4a – 5b)2 – (4c)2
= (m)2 – (4c)2 (Taking 4a – 5b = m)
= (m + 4c)(m – 4c)
= [(4a – 5b) + 4c][(4a – 5b) – 4c] (Putting back m = 4a – 5b)
= (4a – 5b + 4c)(4a – 5b – 4c)
Solution 2(5):
25 – (ab – 3x)2
= (5)2 – (ab – 3x)2
= (5)2 – (m)2 (Taking ab – 3x = m)
= (5 + m)(5 – m)
= [5 + (ab – 3x)][5 – (ab – 3x)] (Putting back m = ab – 3x)
= (5 + ab – 3x)(5 – ab + 3x)
Solution 2(6):
(x + 8)2 – (2x – 3)2
= (m)2 – (n)2 (Taking x + 8 = m and 2x – 3 = n)
= (m + n)(m – n)
= [(x + 8) + (2x – 3)][(x + 8) – (2x – 3)]
(Putting back m = x + 8 and n = 2x – 3)
= (x + 8 + 2x – 3)(x + 8 – 2x + 3)
= (3x + 5)(-x + 11)
Solution 2(7):
121x2 – 22x + 1 – 9a2 – 24ab – 16b2
= (121x2 – 22x + 1) – (9a2 + 24ab + 16b2)
= (11x – 1)2 – (3a + 4b)2
= (m)2 – (n)2 (Taking 11x – 1 = m and 3a + 4b = n)
= (m + n)(m – n)
= [(11x – 1) + (3a + 4b)][(11x – 1) – (3a + 4b)]
(Putting back m = 11x – 1 and n = 3a + 4b)
= (11x – 1 + 3a + 4b)(11x – 1 – 3a – 4b)
= (11x + 3a + 4b – 1)(11x – 3a – 4b – 1)
Solution 2(8):
Solution 2(9):
Solution 2(10):
x4 – 8x2 – 65
= x4 – 13x2 + 5x2 – 65
= x2(x2 – 13) + 5(x2 – 13)
= (x2 – 13)(x2 + 5)
Solution 2(11):
x6 – 27
= (x2)3 – (3)3
= (x2 – 3)[(x2)2 + (x2)(3) + (3)2]
= (x2 – 3)(x4 + 3x2 + 9)
Solution 2(12):
64x3 + 125y3
= (4x)3 + (5y)3
= (4x + 5y)[(4x)2 – (4x)(5y) + (5y)2]
= (4x + 5y)(16x2 – 20xy + 25y2)
Practice 1
Solution 1:
m2 – 16
= (m)2 – (4)2
= (m + 4)(m – 4)
Solution 2:
16x2 – 49y2
= (4x)2 – (7y)2
= (4x + 7y)(4x – 7y)
Solution 3:
4a2b2 – 1
= (2ab)2 – (1)2
= (2ab + 1)(2ab – 1)
Solution 4:
49 – 25x2
= (7)2 – (5x)2
= (7 + 5x)(7 – 5x)
Solution 5:
16x4 – 81y4
= (4x2)2 – (9y2)2
= (4x2 + 9y2)(4x2 – 9y2)
= (4x2 + 9y2) [(2x)2 – (3y)2]
= (4x2 + 9y2)(2x + 3y)(2x – 3y)
Solution 6:
a4b4 – 1
= (a2b2)2 – (1)2
= (a2b2 + 1)(a2b2 – 1)
= (a2b2 + 1) [(ab)2 – (1)2]
= (a2b2 + 1)(ab + 1)(ab – 1)
Solution 7:
x3 – 49x
= x(x2 – 49)
= x[(x)2 – (7)2]
= x(x + 7)(x – 7)
Solution 8:
18x3y3 – 2xy
= 2xy(9x2y2 – 1)
= 2xy [(3xy)2 – (1)2]
= 2xy(3xy + 1)(3xy – 1)
Practice 2
Solution 1(1):
(4x + 3y)2 – 49z2
= (m)2 – (7z)2 (Taking 4x + 3y = m)
= (m + 7z)(m – 7z)
= [(4x + 3y) + 7z][(4x + 3y) – 7z] (Putting back m = 4x + 3y)
= (4x + 3y + 7z)(4x + 3y – 7z)
Solution 1(2):
(ab – 1)2 – 64x2
= (m)2 – (8x)2 (Taking ab – 1 = m)
= (m + 8x)(m – 8x)
= [(ab – 1) + 8x][(ab – 1) – 8x] (Putting back m = ab – 1)
= (ab – 1 + 8x)(ab – 1 – 8x)
= (ab + 8x – 1)(ab – 8x – 1)
Solution 1(3):
81 – (5x – 3y)2
= (9)2 – (m)2 (Taking 5x – 3y = m)
= (9 + m)(9 – m)
= [(9 + (5x – 3y)][9 – (5x – 3y)] (Putting back m = 5x – 3y)
= (9 + 5x – 3y)(9 – 5x + 3y)
= (5x – 3y + 9)(-5x + 3y + 9)
Solution 1(4):
36z2 – (x + 2y)2
= (6z)2 – (m)2 (Taking x + 2y = m)
= (6z + m)(6z – m)
= [6z + (x + 2y)][6z – (x + 2y)] (Putting back m = x + 2y)
= (6z + x + 2y)(6z – x – 2y)
Solution 2(1):
(a + 8)2 – (b – 3)2
= (m)2 – (n)2 (Taking a + 8 = m and b – 3 = n)
= (m + n)(m – n)
= [(a + 8) + (b – 3)][(a + 8) – (b – 3)] (Putting back m = a + 8 and n = b – 3)
= (a + 8 + b – 3)(a + 8 – b + 3)
= (a + b + 5)(a – b + 11)
Solution 2(2):
(3x – 2y)2 – (5a – 3b)2
= (m)2 – (n)2 (Taking 3x – 2y = m and 5a – 3b = n)
= (m + n)(m – n)
= [(3x – 2y) + (5a – 3b)][(3x – 2y) – (5a – 3b)]
(Putting back m = 3x – 2y and n = 5a – 3b)
= (3x – 2y + 5a – 3b)(3x – 2y – 5a + 3b)
Solution 2(3):
(4x + 5)2 – (2y + 3)2
= (m)2 – (n)2 (Taking 4x + 5 = m and 2y + 3 = n)
= (m + n)(m – n)
= [(4x + 5) + (2y + 3)][(4x + 5) – (2y + 3)] (Putting back m = 4x + 5 and n = 2y + 3)
= (4x + 5 + 2y + 3)(4x + 5 – 2y – 3)
= (4x + 2y + 8)(4x – 2y + 2)
= [2(2x + y + 4)][2(2x – y + 1)]
= 4(2x + y + 4)(2x – y + 1)
Solution 2(4):
(ab + 6)2 – (mn – 7)2
= (x)2 – (y)2 (Taking ab + 6 = x and mn – 7 = y)
= (x + y)(x – y)
= [(ab + 6) + (mn – 7)][(ab + 6) – (mn – 7)] (Putting back x = ab + 6 and y = mn – 7)
= (ab + 6 + mn – 7)(ab + 6 – mn + 7)
= (ab + mn – 1)(ab – mn + 13)
Solution 3(1):
36a2 – 12a + 1 – 4b2
= (36a2 – 12a + 1) – (4b2)
= (6a – 1)2 – (2b)2
= (m)2 – (2b)2 (Taking 6a – 1 = m)
= (m + 2b)(m – 2b)
= [(6a – 1) + 2b][(6a – 1) – 2b] (Putting back m = 6a-1)
= (6a – 1 +2b) (6a – 1 – 2b)
= (6a + 2b – 1)(6a – 2b – 1)
Solution 3(2):
64 – x2 – 10x – 25
= (64) – (x2 + 10x + 25)
= (8)2 – (x + 5)2
= (8)2 – (m)2 (Taking x + 5 = m)
= (8 + m)(8 – m)
= [8 + (x + 5)][8 – (x + 5)] (Putting back m = x + 5)
= (8 + x + 5)(8 – x – 5)
= (x + 13)(-x + 3)
Solution 3(3):
m2n2 – 4mn + 4 – x2
= (m2n2 – 4mn + 4) – (x2)
= (mn – 2)2 – (x)2
= (a)2 – (x)2 (Taking mn – 2 = a)
= (a + x)(a – x)
= [(mn – 2) + x][(mn – 2) – x]
(Putting back a = mn – 2)
= (mn – 2 + x)(mn – 2 – x)
= (mn + x – 2)(mn – x – 2)
Solution 3(4):
9m2 – 25x2 + 20xy – 4y2
= (9m2) – (25x2 – 20xy + 4y2)
= (3m)2 – (5x – 2y)2
= (3m)2 – (a)2 (Taking 5x – 2y = a)
= (3m + a)(3m – a)
= [3m + (5x – 2y)][3m – (5x – 2y)] (Putting back a = 5x – 2y)
= (3m + 5x – 2y)(3m – 5x + 2y)
Solution 4(1):
9a2 + 6a + 1 – x2 – 2xy – y2
= (9a2 + 6a + 1) – (x2 + 2xy + y2)
= (3a + 1)2 – (x + y)2
= (m)2 – (n)2 (Taking 3a + 1 = m and x + y = n)
=(m + n)(m – n)
= [(3a + 1) + (x + y)][(3a + 1) – (x + y)] (Putting back m = 3a + 1 and n = x + y)
= (3a + 1 + x + y)(3a + 1 – x – y)
Solution 4(2):
49x2 – 14x + 1 – 64a2 + 16ab – b2
= (49x2 – 14x + 1) – (64a2 – 16ab + b2)
= (7x – 1)2 – (8a – b)2
= (m)2 – (n)2 (Taking 7x – 1 = m and 8a – b = n)
= (m + n)(m – n)
= [(7x – 1) + (8a – b)][(7x – 1) – (8a – b)] (Putting back m = 7x – 1 and n = 8a – b)
= (7x – 1 + 8a – b)(7x – 1 – 8a + b)
= (7x + 8a – b – 1)(7x – 8a + b – 1)
Solution 4(3):
4a2 + 12a + 9 – m2 + 2mn – n2
= (4a2 + 12a + 9) – (m2 – 2mn + n2)
= (2a + 3)2– (m – n)2
= (x)2 – (y)2 (Taking 2a + 3 = x and m – n = y)
= (x + y)(x – y)
= [(2a + 3) + (m – n)][(2a + 3) – (m – n)]
(Putting back x = 2a + 3 and y = m – n)
= (2a + 3 + m – n)(2a + 3 – m + n)
Solution 4(4):
x2 – 14x + 49 – a2 – 12a – 36
= (x2– 14x + 49) – (a2 +12a + 36)
= (x – 7)2– (a + 6)2
= (m)2– (n)2(Taking x – 7 = m and a + 6 = n)
= (m + n)(m – n)
= [(x – 7) + (a + 6)][(x – 7) – (a + 6)]
(Putting back m = x – 7 and n = a + 6)
= (x – 7 + a + 6)(x – 7 – a – 6)
= (x + a – 1)(x – a – 13)
Practice 3
Solution 1:
Solution 2:
Solution 3:
Solution 4:
Practice 4
Solution 1:
Solution 2:
Solution 3:
Solution 4:
Solution 5:
Solution 6:
Practice 5
Solution 1(1):
x3 + 27
= (x)3 + (3)3
= (x + 3)[(x)2 – (x)(3) + (3)2]
= (x + 3)(x2 – 3x + 9)
Solution 1(2):
a3 + 125b3
= (a)3 + (5b)3
= (a + 5b)[(a)2 – (a)(5b) + (5b)2]
= (a + 5b)(a2 – 5ab + 25b2)
Solution 1(3):
8a3b3 – 27
= (2ab)3 – (3)3
= (2ab – 3)[(2ab)2 + (2ab)(3) + (3)2]
= (2ab – 3)(4a2b2 + 6ab + 9)
Solution 1(4):
8x3 – 125
= (2x)3 – (5)3
= (2x – 5)[(2x)2 + (2x)(5) + (5)2]
= (2x – 5)(4x2 + 10x + 25)
Solution 2(1):
a + b = 5
∴ (a + b)2 = (5)2
∴ a2 + 2ab + b2 = 25
∴ a2 + b2 + 2(6) = 25 …..(Putting ab = 6)
∴ a2 + b2 + 12 = 25
∴ a2 + b2 = 25 – 12
∴ a2 + b2 = 13
Now, a3 + b3 = (a + b)(a2 – ab + b2)
= (a + b)(a2 + b2 – ab)
Putting a + b = 5, ab = 6 and a2 + b2 = 13, we have
a3 + b3 = (5)(13 – 6)
= (5)(7)
= 35
Solution 2(2):
a + b = 8, ab = 15 and a2 + b2 = 34
Now,
a3 + b3 = (a + b)(a2 – ab + b2)
= (a + b)(a2 + b2– ab)
= (8)(34 – 15)
= (8)(19)
= 152
Solution 2(3):
a – b = 2, ab = 24 and a2 + b2 = 52
Now,
a3 – b3 = (a – b)(a2 + ab + b2)
= (a – b)(a2 + b2+ ab)
= (2)(52 + 24)
= (2)(76)
= 152
Solution 3(1):
(11)3 + (9)3
= (11 + 9)[(11)2 – (11)(9) + (9)2]
= (20)(121 – 99 + 81)
= (20)(103)
= 2060
Solution 3(2):
(23)3 + (7)3
= (23 + 7)[(23)2 – (23)(7) + (7)2]
= (30)(529 – 161 + 49)
= (30)(417)
= 12510
Solution 3(3):
(45)3 – (25)3
= (45 – 25)[(45)2 + (45)(25) + (25)2]
= (20)(2025 + 1125 + 625)
= (20)(3775)
= 75500
Practice 6
Solution 1:
x2 + 5x + 6
= x2 + 3x + 2x + 6
= x(x + 3) + 2(x + 3)
= (x + 3)(x + 2)
Solution 2:
x2 + 15x + 50
= x2 + 10x + 5x + 50
= x(x + 10) + 5(x + 10)
= (x + 10)(x + 5)
Solution 3:
x2 – 11x + 24
= x2 – 3x – 8x + 24
= x(x – 3) – 8(x – 3)
= (x – 3)(x – 8)
Solution 4:
x2 – 7x +12
= x2 – 4x – 3x + 12
= x(x – 4) – 3(x – 4)
= (x – 4)(x – 3)
Solution 5:
x2 + 6x – 2
= x2 + 9x – 3x – 27
= x(x + 9) – 3(x + 9)
= (x + 9)(x – 3)
Solution 6:
a2 + 4a – 21
= a2 + 7a – 3a – 21
= a(a + 7) – 3(a + 7)
= (a + 7)(a – 3)
Solution 7:
m2 – 2m – 8
= m2 – 4m + 2m – 8
= m(m – 4) + 2(m – 4)
= (m – 4)(m + 2)
Solution 8:
n2 – 4n – 45
= n2 – 9n + 5n – 45
= n(n – 9) + 5(n – 9)
= (n – 9)(n + 5)
Solution 9:
4x2 + 12x + 5
= 4x2 + 10x + 2x + 5
= 2x(2x + 5) + 1(2x + 5)
= (2x + 5)(2x + 1)
Solution 10:
9y4 – 13y2 + 4
= 9y4 – 9y2 – 4y2 + 4
= 9y2(y2 – 1) – 4(y2 – 1)
= (y2 – 1)(9y2 – 4)
= (y2 – 12)[(3y)2 – (2)2]
= (y + 1)(y – 1)(3y + 2)(3y – 2)
Solution 11:
2a2 – 19a – 21
= 2a2 + 2a – 21a – 21
= 2a(a + 1) – 21(a + 1)
= (a + 1)(2a – 21)
Solution 12:
6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (2x – 3)(3x + 1)