GSEB Solutions for Class 8 Mathematics – Factorization – 1 (English Medium)
GSEB SolutionsMathsScience
Exercise
Solution 1:
Solution 2(1):
4ab + 8a – b – 2
= 4a(b + 2) – 1(b + 2)
= (4a – 1)(b + 2)
Solution 2(2):
x2y – 3x2+ y – 3_
= x2(y – 3) + 1(y – 3)
= (x2 + 1)( y – 3)
Solution 2(3):
2x2 – 5a – 5x + 2ax
= 2x2 + 2ax – 5x – 5a
= 2x(x + a) – 5(x + a)
= (2x – 5)(x + a)
Solution 2(4):
3ab + 12 – 4a – 9b
= 3ab – 4a – 9b + 12
= a(3b – 4) – 3(3b – 4)
= (a – 3)(3b – 4)
Solution 2(5):
x2 + 49 + 14x
= x2 + 14x + 49
= (x)2 + 2(x)(7) + (7)2
= (x + 7)2
Solution 2(6):
16a2 + 40ab + 25b2
= (4a)2 + 2(4a)(5b) + (5b)2
= (4a + 5b)2
Solution 2(7):
m4 – 16m2 + 64
= (m2)2 – 2(m2)(8) + (8)2
= (m2 – 8)2
Solution 2(8):
4y3 – 28y2 + 49y = y(4y2 – 28y + 49)
= y[(2y)2 – 2(2y)(7) + (7)2]
= y(2y – 7)2
Solution 2(9):
25x2 + 4y2 + 9z2 + 20xy + 12yz + 30zx
= (5x)2 + (2y)2 + (3z)2 + 2(5x)(2y) + 2(2y)(3z) + 2(3z)(5x)
= (5x + 2y + 3z)2
Solution 2(10):
4m2 + 9n2 + p2 – 12mn + 6np – 4pm
= (2m)2 + (-3n)2 + (-p)2 + 2(2m)(-3n) + 2(-3n)(-p) + 2(-p)(2m)
= (2m – 3n – p)2
Practice 1:
Solution 1:
- 2x2y2 = _2_ × x × x × y × y
- 10a2b = 2 × _5_ × a × _a_ × b
- 6xy = _2_× 3 × x × y
- 15mn2 = 3 × _5_ × m × _n_ × n
Solution 2:
- 25 = 5 × 5
- 6x2y = 2 × 3 × x × x × y
- 20x2y4 = 2 × 2 × 5 × x × x × y × y × y × y
- 24x3y2 = 2 × 2 × 2 × 3 × x × x × x × y × y
- 26xy = 2 × 13 × x × y
- 18a3b = 2 × 3 × 3 × a × a × a × b
Practice 2:
Solution 1:
- x2 – x = x(x – 1)_
x2 – x = _x × x_- x = x(x – 1) - 8x3 + 4x2 = 4x2(2x + 1)
8x3 + 4x2 = 4 × 2 × x × x × x × 4 × x × x = 4x2(2x + 1) - 3a2 – 6 = 3(a2 – 2)
3a2 – 6 = 3 × a × a – 3 × 2 = 3(a2 – 2) - xy – xz = x(y – z)
xy – yz = x × y – x × z = x(y – z)
Solution 2:
- 10x + 5 = 2 × 5 × x + 5
= 5(2x + 1) ….(Taking ‘5’ common) - 5x2 + 15
= 5 × x × x + 3 × 5 = 5(x2 + 3) ….(Taking ‘5’ common) - 7a – 7b
= 7 × a – 7 × b = 7(a – b) ….(Taking ‘7’ common) - -3x + 6
= -3 × x + 2 × 3 = -3(x – 2) ….(Taking ‘-3’ common) - 6x3y2 – 3x
= 2 × 3 × x × x × x × y × y – 3 × x = 3x(2x2y2 – 1) ….(Taking ‘3x’ common) - 9xy2 – 18x2
= 9 × x × y × y – 2 × 9 × x × x = 9x(y2 – 2x) ….(Taking ‘9x’ common) - 8 – 4xy
= 2 × 4 – 4 × x × y = 4(2 – xy) ….(Taking ‘4’ common) - 9x – 27xyz
= 9 × x – 3 × 9 × x × y × z = 9x(1 – 3yz) ….(Taking ‘9x’ common) - 12a2b – 18ab2
= 2 × 6 × a × a × b – 3 × 6× a × b × b = 6ab(2a – 3b) ….(Taking ‘6ab’ common)
Practice 3:
Solution 1:
xy + 2x + 4y + 8
= x(y + 2) + 4(y + 2)
= (x + 4)(y + 2)
Solution 2:
xy – 4x + 3y – 12
= x(y – 4) + 3(y – 4)
= (x + 3)(y – 4)
Solution 3:
= x2y + 5x2 + y + 5
= x2(y + 5) + 1(y + 5)
= (x2 + 1)(y + 5)
Solution 4:
6x2 + 4xy – 3x – 2y
= 2x(3x + 2y) – 1(3x + 2y)
= (2x – 1)(3x + 2y)
Solution 5:
15x – 4a + 6 – 10ax
= 15x + 6 – 10ax – 4a
= 3(5x + 2) – 2a(5x + 2)
= (5x + 2)(3 – 2a)
Solution 6:
10m2n + 9 + 6m + 15mn
= 10m2n + 15mn + 6m + 9
= 5mn(2m + 3) + 3(2m + 3)
= (5mn + 3)(2m + 3)
Practice 4:
Solution 1(1):
Solution 1(2):
Solution 1(3):
Solution 1(4):
First term = 9x2y2 = (3xy)2
Thus, the first term is a perfect positive perfect square.
Last term = 8, which is not a perfect square.
∴ The given polynomial is not a perfect square.
Solution 1(5):
Solution 1(6):
Solution 1(7):
Solution 1(8):
Solution 2(1):
Solution 2(2):
Solution 2(3):
Solution 2(4):
Solution 2(5):
Solution 2(6):
Solution 3:
- x2 + 12x + 36
= (x)2 + 2(x)(6) + (6)2
= (x + 6)2 - 4x2 + 12xy + 9y2
= (2x)2 + 2(2x)(3y) + (3y)2
= (2x + 3y)2 - 9x2 + 48x + 64
= (3x)2 + 2(3x)(8) + (8)2
= (3x + 8)2 - x2 – 8x + 16
= (x)2 – 2(x)(4) + (4)2
= (x – 4)2 - 25x2y2 – 20xy + 4
= (5xy)2 – 2(5xy)(2) + (2)2
= (5xy – 2)2 - 16x2 + 40x + 25
= (4x)2 + 2(4x)(5) + (5)2
= (4x + 5)2 - 81 – 90xy + 25x2y2
= (9)2 – 2(9)(5xy) + (5xy)2
= (9 – 5xy)2 - 3x3 – 30x2 + 75x
= 3x(x2 – 10x + 25)
= 3x[(x)2 – 2(x)(5) + (5)2]
= 3x(x – 5)2
Practice 5:
Solution 1:
9x2 + 4y2 + 1 + 12xy + 4y + 6x
= (3x)2 + (2y)2 + (1)2 + 2(3x)(2y) + 2(2y)(1) + 2(1)(3x)
= (3x + 2y + 1)2
Solution 2:
16a2 + 9b2 + c2 – 24ab + 6bc – 8ca
= (4a)2 + (-3b)2 + (-c)2 + 2(4a)(-3b) + 2(-3b)(-c) + 2(-c)(4a)
= (4a – 3b – c)2
Solution 3:
a4 + 4b2 + 9 + 4a2b – 12b – 6a2
= (a2)2 + (2b)2 + (-3)2 + 2(a2)(2b) + 2(2b)(-3) + 2(-3)(a2)
= (a2 + 2b – 3)2
Solution 4:
9x2 + 16y2 + 25 + 24xy – 40y – 30x
= (3x)2 + (4y)2 + (-5)2 + 2(3x)(4y) + 2(4y)(-5) + 2(-5)(3x)
= (3x + 4y – 5)2
Solution 5:
a2 + 4b2 + c2 – 4ab – 4bc + 2ca
= (a)2 + (-2b)2 + (c)2 + 2(a)(-2b) + 2(-2b)(c) + 2(c)(a)
= (a – 2b + c)2