RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5

Other Exercises

Multiply:

Question 1.
(5x + 3) by (7x + 2)
Solution:
(5x + 3) x (7x + 2)
= 5x (7x + 2) + 3 (7x + 2)
= 35x2 + 10x + 21x + 6
= 35x2 + 31x + 6

Question 2.
(2x + 8) by (x – 3)
Solution:
(2x + 8) x (x – 3)
= 2x (x – 3) + 8 (x – 3)
= 2x2 – 6x + 8x – 24
= 2x2 + 2x – 24

Question 3.
(7x +y) by (x + 5y)
Solution:
(7x + y) x (x + 5y)
= 7x (x + 5y) + y (x + 5y)
= 7x2 + 35xy + xy + 5y2
=7x2 + 36xy + 5y2

Question 4.
(a – 1) by (0.1a2 + 3)
Solution:
(a – 1) x (0.1a2 + 3)
= a (0.1a2 + 3) – 1 (0.1a2+ 3)
= 0.1a3 + 3a-0.1a2-3
= 0.1a3 – 0.1a2 + 3a-3

Question 5.
(3x2 +y2) by (2x2 + 3y2)
Solution:
(3x2+y2) x (2x2 + 3y2)
= 3x2 (2x2 + 3y2) + y2(2x2 + 3y2)
= 6x2 +2 + 9x2y2 + 2x2y2 + 3y2 + 2
= 6x4 + 11 x2y2 + 3y4

Question 6.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 1
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 2

Question 7.
(x6-y6) by (x2+y2)
Solution:
(x6 – y6) x (x2 + y2)
= x6 (x2 + y2) – y6 (x2 + y2)
= x6 x x2 + x6y2 – x2y6 -y6 x y2
= x6 + 2 + x6y2 – x2y6 – y6 +2
= x  + x6y2 – x2y6 – y8

Question 8.
(x2 + y2) by (3a+2b)
Solution:
(x2 + y2) x (3a + 2b)
= x2 (3a + 2b) + y2 (3a + 2b)
= 3x2a + 2x2b + 3y2a + 2y2b
3ax2 + 3av2 + 2bx2 + 2by2

Question 9.
[-3d + (-7ƒ)] by (5d +ƒ)
Solution:
[-3d + (-7ƒ)] x (5d +ƒ)
= -3d x (5d +ƒ) + (-7ƒ) x (5d +ƒ)
= -15d2-3dƒ- 35dƒ- 7ƒ2
= -15d2 – 38dƒ- 7ƒ2

Question 10.
(0.8a – 0.5b) by (1.5a -3b)
Solution:
(0.8a – 0.5b) x (1.5a-3b)
= 0.8a x (1.5a – 36) – 0.56 (1.5a -3b)
= 1.2a2 – 2.4ab – 0.75ab + 1.5b2
= 1.2a2-3.15ab+ 1.5b2

Question 11.
(2x2 y2 – 5xy2) by (x2 -y2)
Solution:
(2x2 y2 – 5xy2) x (x2 -y2)
= 2x2y2 (x2 – y2) – 5x_y2 (x2 – y2)
= 2x2y2 x x2 – 2x2y2 xy2– 5xy2 x x2 + 5x2 xy2
= 2x2 + 2 y2– 2x2 x y2 + 2– 5x1+2 y2+5xy2 + 2
= 2x4y2– 2x2y4 – 5x3y2+ 5xy4

Question 12.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5-q12
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 3

Question 13.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 4
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 5
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 6

Question 14.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 7
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 8

Question 15.
(2x2-1) by (4x3 + 5x2)
Solution:
(2x2-1)x(4x3 + 5x2)
= 2x2 x (4x3 + 5x2) – 1 (4x3 + 5x2)
= 2x2 x 4x3 + 2x2 x 5x2 – 4x3 – 5x2
= 8x2 + 3 + 10x2 + 2-4x3-5x2
= 8x5 + 10x4 – 4x3 – 5x2

Question 16.
(2xy + 3y2) (3y2 – 2)
Solution:
(2xy + 3y2) (3y2 – 2)
= 2xy x (3y2-2) + 3y2 x (3y2-2)
= 2xy x Zy2+ 2xy x (-2) + Zy2 x Zy2 – Zy2 x 2
= 6xy1 + 2– 4xy + 9y2 + 2– 6y2
= 6xy3 – 4xy + 9y4– 6y2
Find the following products and verify the result for x = -1, y = -2 :

Question 17.
(3x-5y)(x+y)
Solution:
(3x-5y)(x+y)
= 3x x (x + y) – 5y x (x + y)
= 3x x x + 3x x y-5y x x-5y x y
= 3x2 + 3xy – 5xy – 5y2
= 3x2 – 2xy – 5y2
Verfification:
x = -1,y = -2
L.H.S. = (3x-5y)(x+y)
= [3 (-1) -5 (-2)] [-1 – 2]
= (-3 + 10) (-3) = 7 x (-3) = -21
R.H.S. = 3x2 – 2xy – 5y2
= 3 (-1)2 – 2 (-1) (-2) -5 (-2)2
=3×1-4-5×4=3-4-20
= 3-24 = -21
∴ L.H.S. = R.H.S.

Question 18.
(x2y-1) (3-2x2y)
Solution:
(x2y-1) (3-2x2y)
= x2y (3 – 2x2y) -1(3-2x2y)
= x2y x 3 – x2y x 2x2y – 1 x 3 + 1 x 2x2y
= 3x2y-2x2 + 2x y1 +1-3 + 2x2y
= 3x2y – 2x4y2– 3 + 2x2y
= 3x2y + 2x2y – 2x4y2 – 3
= 5x2y – 2x4y2 – 3
Verification : (x = -1, y = -2)
L.H.S. = (x2y – 1) (3 – 2x2y)
= [(-1)2 x (-2) -1] [3 – 2 x (-1)2 x (-2)]
= [1 x (-2) -1) [3 – 2 x 1 x (-2)]
= (-2 – 1) (3 + 4) = -3 x 7 = -21
R.H.S. = 5x2y – 2x4y2 – 3
= 5 (-1)2 (-2) -2 (-1)4 (-2)2 -3
5 x 1 (-2) – 2 (1 x 4) -3
= -10-8-3 = -21
∴ L.H.S. = R.H.S

Question 19.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 9
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 10

Simplify :

Question 20.
x2 (x + 2y) (x – 3y)
Solution:
x2 (x + 2y) (x – 3y)
= x2 [x (x – 3y) + 2y (x – 3y)]
= x2 [x2 – 3xy + 2xy – 6y2]
= x2 [x2 – xy – 6y2)
= x2 x x2 – x2 x xy – x26y2
= x4 – x3y – 6x2y2

Question 21.
(x2 – 2y2) (x + 4y)
Solution:
(x2 – 2y2) (x + 4y) x2y2
= [x2 (x + 4y) -2y2 (x + 4y)] x2y2
= (x3 + 4x2y – 2xy2 – 8y3) x2y2
= x2y2 x x3 + x2y2 x 4x2y – 2x2y2 x xy2 – 8x2y2 x y3
= x2 +3 y2 + 4x2 + 2 y2 +1 – 2x2 +1 y2+ 2 – 8x2y2+3
= xy + 44xy3 – 2x3y4 – 8x2y5

Question 22.
a2b2 (a + 2b) (3a + b)
Solution:
a2b2 (a + 2b) (3a + b)
= a2b2 [a (3a + b).+ 2b (3a + b)]
= a2b2 [3a2 + ab + 6ab + 2b2]
= a2b2 [3a2 + lab + 2b2]
= a2b2 x 3a2 + a2b2 x 7ab + a2b2 x 2b2
= 3a2 + 2b2 + 7a2+1 b2+1+ 2a2b2 + 2
= 3a4b2 + 7a3b3 + 2a2b4

Question 23.
x2 (x-y) y2 (x + 2y)
Solution:
x2 (x -y) y2 (x + 2y)
= [x2 x x – x2 x y] [y2 x x + y2 x 2y]
= (x3 – x2y) (xy2 + 2y3)
= x3 (xy2 + 2y3) – x2y (xy2 + 2y3)
= x3 x xy2 + x3 x 2y3 – x2y x xy2 – x2y x 2y3
= x3 +1 y2 + 2x3y3 – x2 +1 y1+ 2 – 2x2y1 + 3
= x4y2 + 2x3y3 – x3y3 – 2x2y4
= x4y2 + x3y3 – 2x2y4

Question 24.
(x3 – 2x2 + 5x-7) (2x-3)
Solution:
(x3 – 2x2 + 5x – 7) (2x – 3)
= (2x – 3) (x3 – 2x2 + 5x – 7)
= 2x (x3 – 2x2 + 5x – 7) -3 (x3 – 2x2 + 5x – 7)
= 2x x x3 – 2x x 2x2 + 2x x 5x – 2x x 7 -3 x x3 – 3 x (-2x2) – 3 x 5x – 3 x (-7)
= 2x4-4x3 + 10x2– 14x-3x3 + 6x2– 15x + 21
= 2x4 – 4x3 – 3x3 + 10x2 + 6x2– 14x- 15x + 21
= 2x4-7x3 + 16x2-29x+ 21

Question 25.
(5x + 3) (x – 1) (3x – 2)
Solution:
(5x + 3) (x – 1) (3x – 2)
= (5x + 3) [x (3x – 2) -1 (3x – 2)]
= (5x + 3) [3x2 – 2x – 3x + 2]
= (5x + 3) [3x2 – 5x + 2]
= 5x (3x2 – 5x + 2) + 3 (3x2 – 5x + 2)
= (5x x 3x2 – 5x x Sx + 5x x 2)+ [3 x 3x2 + 3 x (-5x) + 3×2]
= 15x3 – 25x2 + 10x + 9x2 – 15x + 6
= 15x3 – 25x2 + 9x2 + 10x – 15x + 6
= 15x3 – 16x2 – 5x + 6

Question 26.
(5-x) (6-5x) (2-x)
Solution:
(5-x) (6-5x) (2-x)
= [5 (6 – 5x) -x (6 – 5x)] (2 – x)
= [30 – 2$x – 6x + 5x2] (2 – x)
= (30 – 3 1x + 5x2) (2-x)
= 2 (30 – 31x + 5x2) – x (30 – 31x + 5x2)
= 60 – 62x + 10x2 – 30x + 3 1x2 – 5x3
= 60 – 62x – 30x + 10x2 + 3 1x2 – 5x3
= 60 – 92x + 41x2 – 5x3

Question 27.
(2x2 + 3x – 5) (3x2 – 5x + 4)
Solution:
(2x2 + 3x – 5) (3x2 – 5x + 4)
= 2x2 (3x2 – 5x + 4) + 3x (3x2 – 5x + 4) -5 (3x2 – 5x + 4)
= 2x2 x 3x2 – 2x2 x 5x + 2x2 x 4 + 3x x 3x2 – 3x x 5x + 3x x 4 – 5 x 3x2 – 5 (-5x) -5×4
= 6x4 – 10x3 + 8x2 + 9x3 – 15x2 + 12x – 15x2 + 25x-20
= 6x4 – 10x3 + 9x3 + 8x2 – 15x2 – 15x2 + 12x + 25x – 20
= 6x4 – x3 – 22x2 + 37x – 20

Question 28.
(3x – 2) (2x – 3) + (5x – 3) (x + 1)
Solution:
(3x – 2) (2x – 3) + (5x – 3) (x + 1)
= 3x (2x – 3) -2 (2x – 3) + 5x (x + 1) – 3 (x + 1)
= 6x2 – 9x – 4x + 6 + 5x2 + 5x – 3x – 3
= 6x2 + 5x2 – 9x – 4x + 5x – 3x + 6 – 3
= 11x2– 11x + 3

Question 29.
(5x – 3) (x + 2) – (2x + 5) (4x – 3)
Solution:
(5x – 3) (x + 2) – (2x + 5) (4x – 3)
= [5x (x + 2) -3 (x + 2)] – [2x (4x – 3) + 5 (4x – 3)]
= [5x2 + 1 0x – 3x – 6] – [8x2 – 6x + 20x -15]
= (5x2 + 7x – 6) – (8x2 + 14x – 15)
= 5x2 + lx – 6 – 8x2 – 14x + 15
= 5x2 – 8x2 + 7x – 14x – 6 + 15
= -3x2 – 7x + 9

Question 30.
(3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)
Solution:
(3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)
= [3x (4x + 3y) + 2y (4x + 3y)]-[2x (7x-3y)-y(7x-3y)]
= (12x2 + 9xy + 8xy + 6y2) – (14x2 – 6xy – 7xy + 3y2)
= (12x2 + 17xy + 6y2) – (14x2 – 13xy + 3y2)
= 12x2 + 17xy + 6y2 – 14x2 + 13xy – 3y2
= 12x2 – 14x2 + 17xy + 13xy + 6y2 – 3y2
= -2x2 + 30xy + 3y2
= -2x2 + 3y2 + 30xy

Question 31.
(x2-3x + 2) (5x- 2) – (3x2 + 4x-5) (2x- 1)
Solution:
(x2-3x + 2) (5x- 2) – (3x2 + 4x-5) (2x- 1)
= [5x (x2 – 3x + 2) -2 (x2 – 3x + 2)] – [2x (3x2 + 4x – 5) -1 (3x2 + 4x – 5)]
= [5x3 – 15x2 + 10x – 2x2 + 6x – 4] – [6x3 + 8x2 – 10x – 3x2 – 4x + 5]
= [5x3 – 15x2 – 2x2 + 10xc + 6x – 4] – [6x3 + 8x2 – 3x2 – 10x – 4x + 5]
= (5x3 – 17x2 + 16x-4) – (6x3 + 5x2 – 14x + 5)
= 5x3 – 17x2 + 16x – 4 – 6x3 – 5x2 + 14x – 5
= 5x3 – 6x3 – 17x2 – 5x2 + 16x + 14x – 4 – 5
= -x3 – 22x2 + 30x – 9

Question 32.
x3 – 2x2 + 3x – 4) (x – 1) – (2x – 3) (x2 – x + 1)
Solution:
(x3 – 2x2 + 3x – 4) (x – 1) – (2x – 3) (x2 – x + 1)
= [x (x3 – 2x2 + 3x – 4) – 1 (x3 – 2x2 + 3x – 4)] – [2x (x2 – x + 1) – 3 (x2 – x + 1)]
= [x4 – 2x3 + 3x2 – 4x – x3 + 2x2 – 3x + 4] [2x3 – 2x2 + 2x – 3x2 + 3x – 3]
= (x4 – 2x3 – x3 + 3x2 + 2x2 – 4x – 3x + 4) (2x3 – 2x2 – 3x2 + 2x + 3x – 3)
= (x4 – 3x3 + 5x2 – 7x + 4) – (2x3 – 5x2 + 5x – 3)
= x4 – 3x3 + 5x2 – 7x + 4 – 2x3 + 5x2 – 5x + 3
= x4 – 3x3 – 2x3 + 5x2 + 5x2 – 7x – 5x + 4 + 3
= x4 – 5x3 + 10x2 – 12x + 7

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 are helpful to complete your math homework.

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