# Class 8 RD Sharma Solutions – Chapter 6 Algebraic Expressions And Identities – Exercise 6.5 | Set 2

### Question 17. (3x – 5y) (x + y)

Solution:

To find the product of (3x – 5y) (x + y)    [use distributive property]

â‡’ 3x (x + y) -5y (x + y)

â‡’ 3x2 + 3xy -5xy – 5y2

â‡’ 3x2 -2xy -5y2

The required product is 3x2 -2xy -5y2

Now, putting the values of x = -1 and y = -2 on both the sides and verifying the results .

LHS: (3x – 5y) (x + y)

â‡’ [3(-1) -5(-2)] [(-1) + (-2)]

â‡’ (-3 +10) (-3)

â‡’ (7) (-3)

â‡’ -21

RHS: 3x2 -2xy -5y2

â‡’ 3 (-1)2 -2 (-1)(-2) -5(-2)2

â‡’ 3(1) -2(2) -5(4)

â‡’ 3-4 -20

â‡’ -21

Therefore, LHS = RHS

Hence, result verified.

### Question 18. (x2y – 1) (3 – 2x2y)

Solution:

To find the product  of (x2y – 1) (3 – 2x2y)  [use distributive property]

â‡’ x2y (3 – 2x2y) -1(3 – 2x2y)

â‡’ 3x2y -2x4y2 -3 + 2x2y

â‡’ 5x2y -2x4y2 – 3

The required product is 5x2y -2x4y2 – 3

Now, putting the values of x = -1 and y = -2 on both the sides and verifying the results .

LHS: (x2y – 1) (3 – 2x2y)

â‡’ [(-1)2(-2) – 1] [3 – 2(-1)2(-2)]

â‡’ [(1)(-2) -1] [3 – 2(1)(-2)]

â‡’ (-3)(7)

â‡’ -21

RHS: 5x2y -2x4y2 – 3

â‡’ 5 (-1)2(-2) – 2 (-1)4(-2)2 – 3

â‡’ 5 (-2) -2(4) – 3

â‡’ -10 -8 – 3

â‡’ -21

Therefore, LHS = RHS

Hence, result verified.

### Question 19. (1/3X – Y2/5) (1/3X + Y2/5)

Solution:

To find the product of (1/3X – Y2/5) (1/3X + Y2/5) [use Identity (a-b)(a+b) = a2 – b2]

â‡’ [(1/3X)2 – (Y2/5)2

â‡’ 1/9X2 – Y4 / 25

The required product is  1/9X2 – Y4 / 25

Now, putting the values of x = -1 and y = -2 on both the sides and verifying the results .

LHS: (1/3X – Y2/5) (1/3X + Y2/5)

â‡’ [1/3(-1) – (-2)2/5] [1/3 (-1) + (-2)2 / 5]

â‡’ [-1/3 -4/5] [-1/3 + 4/5]

â‡’ (- 17/15) (7/15)

â‡’ -119/225

RHS: 1/9X2 – Y4/ 25

â‡’ 1/9 (-1)2 – (-2)4/25

â‡’ 1/9 – 16/25

â‡’ -119/225

Therefore, LHS = RHS

Hence, result verified.

### Question 20. x2 (x + 2y) (x -3y)

Solution:

To find the product of  x2 (x + 2y) (x -3y)       [use distributive property]

â‡’ [x2 (x + 2y)] (x -3y)

â‡’ (x3 + 2x2y) (x -3y)

â‡’ x3(x -3y) + 2x2y(x -3y)

â‡’ x4 -3x3y + 2x3y – 6x2y2

â‡’ x4 -x3y -6x2y2

Hence, the required answer is x4 -x3y -6x2y2

### Question 21. (x2 – 2y2) (x + 4y) x2y2

Solution:

To find the product of (x2 -2y2) (x+4y) x2y2     [use distributive property]

â‡’ [x2 (x + 4y) -2y2 (x+4y)] x2y2

â‡’ (x3 + 4x2y -2xy2 -8y3) x2y2

â‡’ x5y2 + 4x4y3 -2x3y4 – 8x2y5

Hence, the required answer is x5y2 + 4x4y3 -2x3y4 -8x2y5

### Question 22. a2b2 (a + 2b) (3a + b)

Solution:

To find the product  of  a2b2 (a+2b) (3a + b)    [use distributive property]

â‡’  a2b [a (3a + b)+2b (3a + b)]

â‡’ a2b2 (3a2 + ab + 6ab + 2b2)

â‡’ 3a4b2 + a3b3 + 6a3b3 + 2a2b4

â‡’ 3a4b2 + 7a3b3 + 2a2b4

Hence, the required answer is 3a4b2 + 7a3b3 + 2a2b4

### Question 23. x2 (x – y) y2 (x+2y)

Solution:

To find the product  of  x2 (x – y) y2 (x+2y)   [use distributive property]

â‡’ x2y2 [x (x+2y) – y (x+2y)]

â‡’ x2y2 (x2 + 2xy – xy -2y2)

â‡’ x4y2 + 2x3y3 – x3y3 -2x2y4

â‡’ x4y2 + x3y3 -2x2y4

Hence, the required answer is x4y2 + x3y3 -2x2y4

### Question 24. (x3 -2x2 + 5x -7) (2x – 3)

Solution:

To find the product  of (x3 -2x2 + 5x -7) (2x – 3)  [use distributive property]

â‡’ 2x (x3 -2x2 + 5x -7) -3 (x3 -2x2 + 5x -7)

â‡’ 2x4 -4x3 + 5x2 – 14x – 3x3 + 6x2 – 15x -21

â‡’ 2x4 – 7x3 +11x2 -29x -21

Hence, the required answer is 2x4 – 7x3 +11x2 -29x -21

### Question 25. (5x + 3) (6 – 5x) (2 – x)

Solution:

To find the product  of  (5x + 3) (6 -5x) (2 – x)   [use distributive property]

â‡’ [5x (6 -5x) + 3 (6 -5x)] (2-x)

â‡’ (30x -25x2 + 18 -15x) (2 – x)

â‡’ (-25x2 -5x +18) (2 – x)

â‡’ 2 (-25x2 -5x +18) -x (-25x2 -5x +18)

â‡’ -50x2 -10x + 36 + 25x3 + 5x2 – 18x

â‡’ 25x3 – 45x2 -28x +36

Hence, the required answer is 25x3 – 45x2 -28x + 36

### Question 26. (5 – x) (6 – 5x) (2 – x)

Solution:

To find the product of (5 – x) (6 – 5x) (2 – x)   [use distributive property]

â‡’ [5(6 – 5x) – x(6 – 5x)] (2 – x)

â‡’ (30 – 25x – 6x + 5x2) (2 – x)

â‡’ (5x2 – 31x + 30) (2 – x)

â‡’ 2(5x2 – 31x + 30) – x(5x2 – 31x + 30)

â‡’ 10x2 – 62x + 60 – 5x3 + 31x2 – 30x

â‡’ -5x3 + 41x2 – 92x + 60

Hence, the required answer is -5x3 + 41x2 – 92x + 60

### Question 27. (2x2 + 3x – 5) (3x2 -5x + 4)

Solution:

To find the product of (2x2 + 3x – 5) (3x2 – 5x + 4) [use distributive property]

â‡’ [2x2 (3x2 – 5x + 4) + 3x(3x2 – 5x + 4) – 5(3x2 – 5x + 4)]

â‡’ (6x4 – 10x3 + 8x2) + (9x3 – 15x2 + 12x) + (- 15x2 + 25x – 20)

â‡’ 6x4 – x3 – 22x2 + 37x – 20

Hence, the required answer is 6x4 – x3 – 22x2 + 37x – 20

### Question 28. (3x – 2) (2x – 3) + (5x – 3) (x + 1)

Solution:

To find the product of (3x – 2) (2x – 3) + (5x – 3) (x + 1) [use distributive property]

â‡’ [3x (2x – 3) – 2(2x – 3)] + [5x(x + 1) -3(x + 1)]

â‡’ (6x2 – 9x – 4x + 6) + (5x2 + 5x – 3x – 3)

â‡’ 11x2 – 11x + 3

Hence, the required answer is 11x2 – 11x + 3

### Question 29. (5x – 3)(x + 2) – (2x + 5) (4x – 3)

Solution:

To find the product of (5x – 3)(x + 2) – (2x + 5) (4x – 3) [use distributive property]

â‡’ [5x (x + 2) – 3 (x + 2)] – [2x (4x – 3) + 5(4x – 3)]

â‡’ (5x2 + 10 x – 3x – 6) – (8x2 – 6x + 20x -15)

â‡’ (5x2 + 7x – 6) – (8x2 + 14x – 15)

â‡’ -3x2 – 7x + 9

Hence, the required answer is -3x2 – 7x + 9

### Question 30. (3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)

Solution:

To find the product of (3x + 2y) (4x + 3y) – (2x – y) (7x – 3y) [use distributive property]

â‡’ [3x (4x + 3y) + 2y (4x + 3y)] – [2x(7x – 3y) – y(7x – 3y)]

â‡’ (12x2 + 9xy + 8xy + 6y2) – (14x2 – 6xy – 7xy + 3y2)

â‡’ (12x2 + 17xy + 6y2) – (14x2 – 13xy + 3y2

â‡’ -2x2 + 30xy + 3y2

Hence, the required answer is -2x2 + 30xy + 3y2

### Question 31. (x2 – 3x + 2) (5x – 2) – (3x2 + 4x – 5)(2x – 1)

Solution:

To find the product of (x2 – 3x + 2) (5x – 2) – (3x2 + 4x – 5)(2x – 1)  [use distributive property]

â‡’ [5x (x2 – 3x + 2) – 2(x2 – 3x + 2)] – [2x (3x2 + 4x – 5) – 1(3x2 + 4x – 5)]

â‡’ (5x3 -15x2 + 10x -2x2 + 6x -4) – (6x3 + 8x2 -10x – 3x2 – 4x + 5)

â‡’ (5x3 -17x2 +16x – 4) – (6x3 + 5x2 – 14x + 5)

â‡’ -x3 -22x2 + 30x – 9

Hence, the required answer is -x3 – 22x2 + 30x – 9

### Question 32. (x3 – 2x2 + 3x – 4)(x -1) – (2x – 1) (x2 – x + 1)

Solution:

To find the product of (x3 – 2x2 + 3x – 4)(x -1) – (2x – 1) (x2 – x + 1) [use distributive property]

â‡’ [x(x3 -2x2 +3x -4) – 1(x3 – 2x2 + 3x – 4)] – [2x (x2 – x + 1) – 1(x2 – x + 1)]

â‡’ (x4 -2x3 + 3x2 -4x -x3 + 2x2 – 3x + 4) – (2x3 – 2x2 + 2x – x2 + x – 1)

â‡’ (x4 -3x3 + 5x2 -7x + 4) – (2x3 -3x2 + 3x -1)

â‡’ x4 – 5x3 + 8x2 – 10x + 5

Hence, the required answer is x4 – 5x3 + 8x2 – 10x + 5

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