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Brilliant’s Composite Mathematics Class 8 Solutions Chapter 4 Algebraic Expressions
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Brilliant’s Composite Mathematics Class 8 Math Book, Chapter 4, Algebraic Expressions. Here students can easily find step by step solutions of all the problems for Algebraic Expressions, Exercise 4.1, 4.2 and 4.3 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
Algebraic Expressions Exercise 4.1 all Questions Solution
Question no – (1)
Solution :
(i) (x + 2)2
= (x)2 + 2.(x).(2) + (2)2
= x2 + 4x + 4
(ii) (x + 1/x)3
= (x)3 + 3.(x)2.(1/x) + 3(x) 91/x)2 + (1/x)3
= x3 + 3x + 3/x + 1/x3
(iii) (2x – 1/2)2
= (2x)2 – 2(2x).(1/2) + (1/2)2
= 4x2 – 2x + 1/4
(iv) (x – 2)(x – 1/2)
= x2 + {(-2) + (- 1/2)}x + (- 2) × (- 1/2)
= x2 – 5/2x + 1
(v) (z – 7)(z + 5)
= z2 + {(-7) + (5)}z – (-7) × (5)
= z2 – 2z + 35
(vii) (5x + y + z)2
= (5x)2 + (y)2 + (z)2 + 2(5x)y + 2yz + 2(5x)z
= 25x2 + y2 + z2 + 10xy + 2xy + 10zx
(viii) (3x – p/2 + 3q)2
= (3x)2 + (p/2)2 + (3q)2 – 2 (3x)(p/2) – 2.(p/2)(3q) + 2.(3x)(3q)
= 9x2 + p2/4 + 9q2 – 3px – 3pq + 18qx
(ix) (4 – y)3
= (4)3 – 3.(4)2 y + 3(4) (y2) – (y)3
= 64 – 48y + 12y2 – y3
(xi) (2/3x – 5/3z)3
= (2/3x)3 – 3.(2x/3)2 (5z/3) + 3 (2x/3)(5z/3)2 – (5/3 – z)3
= 8×3/27 – 20x2z/9 + 50xz2/9 – 125z3/27
(xii) (0.9x + 0.7y)(0.81×2 – 0.63xy + 0.49y2)
= (0.9x)3 + (0.7y)3
= 0.729×3 + 0.343y3
(xvi) (3x – 2y – z)2
= (3x)2 + (2y)2 + 9z)2 – 2 (3x)(2y) + 2(2y)(z) – 2(3x)(z)
= 9x2 + 4y2 + z2 – 12xy + 4yz – 6zx
(xvii) (2x + 5)(4x2 – 10x + 25)
= (2x + 5){(2x)2 – (2x).5 + (5)2]
= (2x)3 + (5)3
= 8x3 + 125
Question no – (2)
Solution :
(i) 107 × 103
= (105 + 2) (105 – 2)
= (105)2 – (2)2
= 11025 – 4
= 11021
(ii) (5001/2)2
= (500+ 1/2)2
= (500)2 + 2.500 1/2 + (1/2)2
= 250000 + 500 + 1/4
= 250500 1/4
(iii) (99.33)2
= (9933/100)2
= (9900 + 33/100)2
= (9900)2 + 2(9900)(33) + (33)2/(100)2
= 98010000 + 653400 + 1089/10000
(vi) (497)3
= (500 – 3)3
= (500)3 – 3(500)(3)2 – (3)3
= 125000000 – 2250000 + 13500 – 27
= 12276373
(viii) (1001)3
= (1000)3 + 3(1000)2 (1) + 3 (1000) (12) + (1)3
= 1000000000 + 3000000 + 3000 + 1
= 1003003001
(xii) (2x + 4y)(2x – 4y)
= (2x)2 – (4y)2
= 4x2 – 16y2
(xv) (x2 – 1)(x2 + 1)
= (x2)2 – (1)2
= x4 -1
Question no – (3)
Solution :
(i) (2x + p – c)2 – (2x – p + c)2
= {(2x)2 + (p)2 + (c2) + 2(2x)(p) – 2(p)(c)} – {(2x)2 + p2 + c2 – 2(2x)(p) – 2 (p)(c) + 2 (2x)(c)}
= {4x2 + p2 + c2 + 4px – 2pc – 4cx} – {4x2 + p2 + c2 – 4px – 2pc + 4cx}
= 8px – 8cx
(ii) (2x + 5)3 – (2x – 5)3
= {(2x)3 + 3.(2x)2.5 + 3(2x)(5)2 + (5)3} – {(2x)3 – 3(2x)2(5) + 3(2x)(5)2 – (5)3}
= {8x3 + 60x2 + 150x + 125} – {8x3 – 60x2 + 150x – 125}
= 120x2 + 250
(iii) (x/3 + y/5)3 – (x/3 – y/5)3
= {(x/3)3 + 3 (x/3)2(y/5) + 3 (x/3)(y/5)2 + (y/5)3} – {(x/3)3 – 3(x/3)2(y/5) + 3 (x/3)(y/5)2 – (y/5)3}
= {x3/27 + 3x2y/15 + 3y2/5 + y3/125} – {x3/27 – x2y/15 + xy2/5 – y3/125}
= 2x2y/15 + 2y3/125
(iv) (2x – z)3 + (2x + z)3
= {8x3 – 12x2z + 6xz2 – z3} + {8x3 + 12x2z + 6xz2 + z3}
= 16x3 + 12xz2
Question no – (4)
Solution :
(i) x2 + y2 and (x – y)2, when x + y = 6, xy = 2.
∴ x2 + y2
= (x + y)2 – 2xy
= (6)2 – 2(2)
= 36 – 4
= 32
∴ (x – y)2
= (x + y)2 – 4xy
=(6)2 – 4(2)
= 36 – 4
= 28
(ii) x2 + y2 and (x + y)2 when x – y = 10, xy = 52
∴ (x2 – y2)
= (x – y)2 + 2xy
= (10)2 + 2(52)
= 100 + 104
= 204
∴ (x + y)2
= (x – y)2 + 4xy
= (10)2 + 4(52)
= 100 + 208
= 308
(iii) x4 + 1/x4 and (x2 – 1/x2)2, when x2 + 1/x2 = 5 and x × 1/x = 2
∴ x4 + 1/x4
= x2 + 1/x2)2 – 2.x2.1/x2
= (5)2 – 2
= 25 – 2
= 23
∴(x2 – 1/x2)2
= (x2 + 1/x2)2 – 4.x2.1/x2
= (5)2 – 4
= 25 – 4
= 21
(iv) a3 + 8b3 if a + 2b = 10 and ab = 15
∴ a3 + 8b3
= (a + 2b)(a2 – 2ab + 4b2)
= (a + 2b){(a2 + 4b2) – 2ab}
= (a + 2b){(a + 2b)2 – 4ab – 2ab}
= (a + 2b){(a + 2b)2 – 6ab}
= 10 {(10)2 – 6(15)}
= 10{100 – 90}
= 10 × 10
= 100
(v) x3 – y3 if x – y = – 8 and xy = – 12
∴ x3 – y3
= (x – y)3 + 3xy(x – y)
= (-8)3 + 3(- 12)(- 8)
= – 64 + 288
= 224
(vii) x3 + 1/x3, if x – 1/x = √5
Firstly, (x + 1/x)2 = (x – 1/x)2 + 4.x.1/x
Or, (x + 1/x)2 = (√5)2 + 4 = 5 + 4 = 9
Or, x + 1/x = √9 = 3
(x3 + 1/x7) = (x + 1/x)3 + 3x.1/x(x + 1/x)
= (3)3 + 3(3)
= 27 + 9
= 36
(viii) x3 – 1/x3, if x – 1/x = 10
∴ x3 – 1/x3
= (x – 1/x)3 + 3x.1/x(x – 1/x)
= (10)3 – 3(10)
= 1000 – 30
= 970
(ix) x2 – 1/x2 and (x – 1/x)2, when x + 1/x = 8
∴ x2 + 1/x2
= (x + 1/x)2 – 2.x.1/x
= (8)2 – 2
= 64 – 2
= 62
(x) ab when a – b = 2 and a2 + b2 = 36
∴ a2 + b2
= (a – b)2 + 2ab
= (2)2 + 2ab
= 36
Or, 4 + 2ab = 36
Or, 2ab = 36 – 4
= 32
Or, 2ab = 32
Or, ab = 32/2
= 16
Question no – (5)
Solution :
(i) (3x – 2y – 2)2 = __ x2 + __ y2 + __ z2 – xy + __ yz – __ zx.
∴ (3x – 2y – 2)2 = 9 x2 + 4 y2 + 0 y2 – xy + 0 yz – 0 zx .
(ii) (p – yz)3 = __ p3 – __ p2yz + __ y2z2 – __ y3z3
∴ (p – yz)3 = 1 p3 – 3 p2yz + 3 y2z2 – 1 y3z3
(iii) ( __- 2/3)3 = 64/125 y3 – __ y2 + __ y – __
∴ (4y/5 – 2/3)3 = 64/125 y3 – 48/25 y2 + 29/45 y – 8/27
(iv) (x + __ y) × (x2 – 3xy + 9y2) = x3 + __ y3
∴ (x + 3y) × (x2 – 3xy + 9y2) = x3 + 27y3
(v) (x + __ )(x2 – __ + 1) = x3 + 1
∴ (x + 1)(x2 – x + 1) = x3 + 1
Question no – (7)
Solution :
(i) 453 – 653 + 203
= (453 + 203) – 653
= (45 + 20)3 – 3 × 45 × 20(45 + 20) – 653
= 653 – (2700 × 65) – 653
= – 55500
(ii) (9.8)3 – (11.3)3 + (1.5)3
= (9.83 + 91.5)3 – (11.3)3
= (9.8 + 1.5)3 – 3 × 9.8 ×1.5 (9.8 + 1.5) – (11.3)3
= (11.3)3 – 498.33 – (11.3)3
= – 498.33
(iii) (16)3 – (41)3 + (25)3
= (16)3 + (25)3 – (41)3
= (16 + 25)3 – 3 × 16 × 25(25 + 16).(41)
= (41)3 – 1200 × 41 – (41)3
= – 49200
Algebraic Expressions Exercise 4.2 all Questions Solution
Question no – (1)
Solution :
x2 – 2x – 15
= x2 –(5 – 3)x – 15
= x2 – 5x + 2x – 15
= x(x – 5) + 2(x – 5)
= (x – 5)(x + 2)
Question no – (2)
Solution :
x2 – 7x + 12
= x2 – (4 + 3)x + 12
= x2 – 4x – 3x + 12
= x (x – 4) – 3(x – 4)
= (x – 4)(x – 3)
Question no – (3)
Solution :
x2 + x – 12
= x2 + (4 – 3)x – 12
= x2 + 4x – 3x -12
= x(x + 4) – 3 (x + 4)
= (x + 4)(x – 3)
Question no – (5)
Solution :
x2 + 2x – 15
= x2 + (5 – 3)x – 15
= x2 + 5x – 3x – 15
= x(x + 5) – 3 (x + 5)
= (x + 5) (x – 3)
Question no – (6)
Solution :
x2 + 7x + 10
= x2 + (5 + 2)x + 10
= x2 + 5x + 2x + 10
= x(x + 5) + 2 (x + 5)
= (x + 5) (x + 2)
Question no – (7)
Solution :
x2 + 10x + 25
= x2 + (5 + 5)x + 25
= x2 + 5x + 5x + 25
= x(x + 5) + 5(x + 5)
= (x + 5) (x + 5)
Question no – (9)
Solution :
x2 – 3xy + 2y2
= x2 – (2y + y)x + 2y2
= x2 – 2yx – xy + 2y2
= x(x – 2y) – y (x – 2y)
= (x – 2y)(x – y)
Question no – (10)
Solution :
x2 – 9x + 18
= x2 – (6 + 3)x + 18
= x2 – 6x – 3x – 3x + 18
= x(x – 6) – 3 (x – 6)
= (x – 6)(x – 3)
Question no – (11)
Solution :
12x2 – 23xy + 10y2
= 12x2 – (15 + 8)xy + 10y2
= 12x2 – 15xy – 8xy + 10y2
= 3x(4x – 5y) – 2y(4x – 5y)
= (4x – 5y)(3x – 2y)
Question no – (13)
Solution :
1 + 14x2 + 49x4
= 1 + (7 + 7) x2 + 49x4
= 1 + 7x2 + 7×2 + 49x4
= (1 + 7x2) + 7x2 (1 + 7x2)
= (1 + 7x2) (1 + 7x2)
Question no – (14)
Solution :
25x2 – 20x + 4
= 25x2 – (10 + 10)x + 4 – 25x2 – 10x – 10x + 4
= 5x(5x – 2) -2(5x – 2)
= (5x – 2) (5x – 2)
Question no – (15)
Solution :
36x4 – 60x2y + 25y2
= 36x4 – (30 + 30)x2y + 25y2
= 36x4 – 30x2y – 30x2y + 25y2
= 6×2 (6x2 – 5y) – 5y (6x2 – 5y)
= (6x2 – 5y)(6x2 – 5y)
Question no – (16)
Solution :
4x2 – 16x + 7
= 4x2 – (14 + 2)x + 7
= 4x2 – 14x – 2x + 7
= 2x(2x – 7) – 1 (2x – 7)
= (2x – 7)(2x – 1)
Question no – (17)
Solution :
6x2 – x – 2
= 6x2 – (4 – 3)x –2
= 6x2 – 4x + 3x – 2
= 2x (3x – 2) + 1 (3x – 2)
= (3x – 2) (2x + 1)
Question no – (18)
Solution :
12y2 + 28y – 5
= 12y2 + (30 – 2)y – 5
= 12y2 + 30y – 2y – 5
= 6y (2y + 5) – 1(2y + 5)
= (2y + 5)(6y – 1)
Question no – (19)
Solution :
5x2 – 23xy – 10y2
= 5x2 – (25 – 2)xy – 10y2
= 5x2 – 25xy + 2x – 10y2
= 5x (5x – 5y) + 2y(x – 5y)
= (x – 3y)(5x + 2y)
Question no – (20)
Solution :
10p2 + 11p + 3
= 10p2 + (6 + 5) p + 3
= 10p2 + 6p + 5p + 3
= 2p (5p + 3) + 1 (5p + 3)
= (5p + 3)(2p + 1)
Question no – (21)
Solution :
12x2 + 7xy – 10y2
= 12x2 + (15 – 8)xy – 10y2
= 12x2 + 15xy – 8xy – 10y2
= 3x (4x + 5y) – 2y(4x + 5y)
Question no – (22)
Solution :
6x2 + 35xy – 6y2
= 6x2 + (36 – 41) xy 2
= 6x2 + 35xy – xy – 6y2
= 6x (x + 6y) – y (x + 6y)
= (x + 6y)(x – y)
Question no – (23)
Solution :
x2 – 3√3x + 6
= x2 – (2√3 + √3)x +6
= x2 – 2√3x – √3x + 6
= x (x – 2√3) – √3 (x – 2√3)
= (x – 2√3)(x – √3)
Question no – (24)
Solution :
15x2 – 16xyz – 15y2z2
= 15x2 – (25 – 9)xyz – 15y2z2
= 15x2 – 25xyz + 9xyz – 15y2z2
= 5x (3x – 5yz) + 3yz(3x – 5yz)
= (3x – 5yz)(5x + 3yz)
Question no – (25)
Solution :
14x2 + 11xy – 15y2
= 14x2 + (21 – 10)xy – 15y2
= 14x2 + 21xy – 10xy – 10xy – 15y2
= 7x(2x + 3y) – 5y(2x + 3y)
= (2x + 3y) (7x – 5y)
Question no – (26)
Solution :
x2 + 2√3x – 24
= x2 + (4√3 – 2√3)x- 24
= x2 + 4√3x – 2√3x – 24
= x(x + 4√3) – 2√3(x + 4√3)
= (x + 4√3)(x – 2√3)
Question no – (27)
Solution :
x2 + 3√2x + 4
= x2 + (2√2 + √2)x + 4
= x2 + 2√2x + √2x + 4
= x (x + 2√2) + 2 (x + 2√2)
= (x + 2√2)(x + √2)
Question no – (29)
Solution :
2 – a – a2
= 2 – (2 – 1)a – a2
= 2 – 2a + a – a2
= 2(1 – a) + a (1 – a)
= (1 – a)(2 + a)
Question no – (30)
Solution :
6a2 + 27ab – 3b2
= 6a2 + (18 – 1)ab – 3b2
= 6a2 + 18ab – ab – 3b2
= 6a (a + 3b) – b (a + 3b)
= (a + 3b) (6a – b)
Algebraic Expressions Exercise 4.3 all Questions Solution
Question no – (1)
Solution :
(i) 128x3 + 54z3
= 2 (64x3 + 27z3)
= 2 (4x + 3z)(16x2 – 12xz + 9z2)
(ii) 1 – 8p3
= (1)3 – (2p)3
= (1 – 2p)(1 + 2p + 4p2)
(iii) 54p3q + 2p3q4
= 2pq3 (27 + q3)
= 2p3q(3 + q)(9 – 3q + q2)
(iv) y3 – 125
= (y)3 – (5)3
= (y – 5)(y2 + 5y + 25)
(v) 27x3 + 64y3
= (3x)3 (4y)3
= (3x + 4y) (9x2 – 12xy + 16y2)
(vi) 3√3 x3 – 125
= (√3x)3 – (5)3
= (√3x + 5)(3x2 + 5√3x + 25)
(vii) 8 + 125x3
= (2)3 + (5x)3
= (2 + 5x)(4 – 10x + 25x2)
(viii) 512x3 – 64y3
= 64[(8x3) – y3]
= 64 (2x – y)(4x2 + 2xy + y2)
(ix) 64 – 343y3
= (4)3 – (7y)3
= (4 – 7y)(16 + 28y + 49y2)
(x) 0.001x3 – 0.125y3
= (0.1x)3 – (0.5y)3
= (0.1x – 0.5y)(0.01x2 + 0.05x + 0.25y2)
(xi) 1/216 x3 – 1/125 y3
= (x/6)3 – (y/5)3
= (x/6 – y/5)(x2/36 + xy/30 + y2/25)
(xii) y3 + 125
= (y)3 + (5)3
= (y + 5)(y2 – 5y + 25)
(xiii) 1000x3 + y3
= (10x)3 + (y)3
= (10x + y)(100x2 – 10xy + y2)
(xiv) (2x – 3)3 – y3
= (2x – 3 – y){(2x – 3)2 + y (2x – 3) + y2}
(xv) x6 – y6
= (x3)2 – (y3)2
= (x3 + y3)(x3 – y3)
= (x + y)(x2 – xy + y2)(x – y)(x2 + xy + y2)
Question no – (2)
Solution :
(i) 4x2 + y2 + 9z2 + 4xy – 6yz – 12zx
= (2x)2 + y2 + (3z)2 + 2.(2x).y – 2.y(3z) – 2 (2x)(3z)
= (-2x – y + 3z)2
(ii) x2 + xy + y2/4 + 1 + 2x + y
= (x)2 + (4/2)2 + (1)2 + 2.x.y/2 + 2.y/2 .1 + 2. x .1
= (x + y/2 + 1)2
(iii) 9a2 + b2 + 4c2 – 6ab – 4bc + 12ac
= (3a)2 + (b)2 + (2c)2 – 2 (3a)(b) – 2 (b)(2c) + 2(3a)(2c)
= (- 3a – b + 2c)2
(iv) x2 + 25y2 + 16z2 – 10xy + 40yz – 8xz
= (x)2 + (5y)2 + (4z)2 – 29x)(5y) + 2 (5y)(4z) – 2 (x) (4z)
= (x – 5y – 4z)2
(v) x2 + 4y2 + z2 + 4xy + 4yz + 2zx
= (x)2 + (2y)2 + (z)2 + 2(x)(2y) + 2(2y)(z) + 2(z)(x)
= (x + 2y + z)2
Question no – (3)
Solution :
(i) 27x3 + 8z3 + 54x2z + 36xz2
= (3x)3 + (2z)3 + 3(3x)(2z)[(3x) + (2z)]
= (3x + 2z)3
(ii) x3 + 64y3 + 12x2y + 48xy2
= (x)3 + 94y)3 + 3(x)(4y)[x + 4y]
= (x + 4y)3
(iii) x3/8 – 64 – 3x2 + 24x
= (x/2)3 – (4)3 – 3(x/2)2(4) + 3(x/2) (42)
= (2x – 3y)3
(iv) 8x3 – 27y3 – 36x2y + 54xy2
= (2x)3 – (3y)3 – 3 (2x)2 )(3y) + 3(2x) (3y)2
= (3x – y)3
(v) 27x3 – y3 – 27x2y + 9xy2
= (3x)3 – (y)3 – 3 (3x)2 9y) + 3 (3x) (y2)
= (3x – y)3
(vi) a3/8 – b3/64 – 3a2b/16 + 3ab2/32
= (a/2)3 – (b/4)3– 3 (a/2)2 (b/4) + 3(a/2)(b/4)2
= (a/2 – b/4)3
Question no – (4)
Solution :
(i) 8x3 + 27y3 + z3 – 18xyz
= (2x)3 + (3y)3 + (z)3 – 3 (2x) (3y) (z)
= (2x + 3y + z) (4×2 + 9y2 + z – 6xy – 3yz – 2xz)
(ii) l3 + m3 – n3 + 3lmn
= (l)3 + (m)3 + (- n)3 -3lm(- n)
= (l + m – n)(l2 + m2 + n2 –lm + mn + nl)
(iii) 125x3 – y3 + 8z3 + 30xtz
= (5x)3 + (-y)3 + (2z)3 – 3 (5x)(- y)(2z)
= (5x – y + 2z)(25x2 + y2 + 4z2 + 5xy + 2yz – 10zx)
(iv) – 64a3 + b3 – 27c3 – 36abc
= (- 4)3 + (b)3 + (- 3c)3 – 3(- 4a)(b)(- 3c)
= (- 4a + b – 3c)(16a2 + b2 + 9c2 + 4ba + 3bc – 12ac)
(v) 3x6 + 24y3 + 81z9 – 54x2yz3
= 3[(x2)3 + (2y)3 + (3z3)3 – 3(x2)(2y)(3z3)]
= 3(x2 + 2y + 3z3)(x4 + 4y2 + 9z6 – 2x2y – 6yz3 – 3x2z3)
Question no – (5)
Solution :
(i) (x – y)3 + (y – z)3 + (z – x)3
∴ 3(x – y)(y – z)(z – x)
Since, (x – y) + (y – z) + (z – x)
= x – y + y – z + z – x
= 0
(ii) (3a – 5b)3 + (5b – 9c)3 + (9c – 3a)3
Since, (3a – 5b) + (5b – 9c) + (9c – 3a)
= 3a – 5b + 5b – 9c + 9c – 3a
= 0
∴ 3 (3a – 5b)(5b – 9c)(9c – 3a)
(iii) (a – b + c)3 + (b – c + a)3 + (- 2a)3
Since, (a – b + c) + (b – c + a) + (- 2a) = 0
Therefore, (a – b + c) + (b – c + a)3 + ( – 2a)3
= 3 (a – b + c)(b – c + a)(- 2a)
(iv) (x – 3y)3 + (3y – 5z)3 + (5z – x)3
Since, (x – 3y) + (3y – 5z) + (5z – x) = 0
Therefore, (x – 3y)3 + (3y – 5z03 + (5z – x)3
= 3(x – 3x)(3y – 5z)(5z – x)
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