RD Sharma Solutions for Class 8 Chapter 6 - Algebraic Expressions and Identities Exercise 6.4

RD Sharma Class 8 Solutions Chapter 6 Ex 6.4 PDF Free Download

Students can avail free pdf of RD Sharma Solutions for Class 8 Maths Exercise 6.4 of Chapter 6 Algebraic Expressions and Identities which are provided here. In Exercise 6.4 of Chapter 6 algebraic expressions and identities, we shall discuss problems based on the multiplication of a monomial and a binomial. Our expert tutors at BYJU’S have solved the questions from the RD Sharma textbook in a step by step manner for easy understanding of the concepts, which helps students increase their speed in solving the questions quickly in their exams. RD Sharma Class 8 Solutions pdf can be downloaded easily from the links provided below.

Download the pdf of RD Sharma For Class 8 Maths Exercise 6.4 Chapter 6 Algebraic Expressions and Identities

 

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rd sharma class 8 maths chapter 6
rd sharma class 8 maths chapter 6
rd sharma class 8 maths chapter 6
rd sharma class 8 maths chapter 6
rd sharma class 8 maths chapter 6
rd sharma class 8 maths chapter 6
rd sharma class 8 maths chapter 6

 

Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 6.4 Chapter 6 Algebraic Expressions and Identities

Find the following products:

1. 2a3 (3a + 5b)

Solution:

Let us simplify the given expression

2a3 (3a + 5b)

(2a3 × 3a) + (2a3 × 5b)

6a3+1 + 10a3b

6a4 + 10a3b

2. -11a (3a + 2b)

Solution:

Let us simplify the given expression

-11a (3a + 2b)

(-11a × 3a) + (-11a × 2b)

-33a2 – 22ab

3. -5a (7a – 2b)

Solution:

Let us simplify the given expression

-5a (7a – 2b)

(-5a × 7a) – (-5a × 2b)

-35a2 + 10ab

4. -11y2 (3y + 7)

Solution:

Let us simplify the given expression

-11y2 (3y + 7)

(-11y2 × 3y) + (-11y2 × 7)

-33y3 – 77y2

5. 6x/5(x3 + y3)

Solution:

Let us simplify the given expression

6/5x (x3 + y3)

(6/5x × x3) + (6/5x × y3)

6/5x4 + 6/5xy3

6. xy (x3 – y3)

Solution:

Let us simplify the given expression

xy (x3 – y3)

(xy × x3) – (xy × y3)

x4y – xy4

7. 0.1y (0.1x5 + 0.1y)

Solution:

Let us simplify the given expression

0.1y (0.1x5 + 0.1y)

(0.1y × 0.1x5) + (0.1y × 0.1y)

0.01x5y + 0.01y2

8. (-7/4ab2c – 6/25a2c2) (-50a2b2c2)

Solution:

Let us simplify the given expression

(-7/4ab2c – 6/25a2c2) (-50a2b2c2)

(-7/4ab2c × -50a2b2c2) – (6/25a2c2 × -50a2b2 × c2)

350/4a3b4c3 + 12a4b2c4

175/2a3b4c3 + 12a4b2c4

9. -8/27xyz (3/2xyz2 – 9/4xy2z3)

Solution:

Let us simplify the given expression

-8/27xyz (3/2xyz2 – 9/4xy2z3)

(-8/27xyz × 3/2xyz2) – (-8/27xyz × 9/4xy2z3)

-4/9x2y2z3 + 2/3x2y3z4

10. -4/27xyz (9/2x2yz – 3/4xyz2)

Solution:

Let us simplify the given expression

-4/27xyz (9/2x2yz – 3/4xyz2)

(-4/27xyz × 9/2x2yz) – (-4/27xyz × 3/4xyz2)

-2/3x3y2z2 + 1/9x2y2z3

11. 1.5x (10x2y – 100xy2)

Solution:

Let us simplify the given expression

1.5x (10x2y – 100xy2)

(1.5x 10x2y) – (1.5x × 100xy2)

15x3y – 150x2y2

12. 4.1xy (1.1x – y)

Solution:

Let us simplify the given expression

4.1xy (1.1x – y)

(4.1xy × 1.1x) – (4.1xy × y)

4.51x2y – 4.1xy2

13. 250.5xy (xz + y/10)

Solution:

Let us simplify the given expression

250.5xy (xz + y/10)

(250.5xy × xz) + (250.5xy × y/10)

250.5x2yz + 25.05xy2

14. 7/5x2y (3/5xy2 + 2/5x)

Solution:

Let us simplify the given expression

7/5x2y (3/5xy2 + 2/5x)

(7/5x2y × 3/5xy2) + (7/5x2y × 2/5x)

21/25x3y3 + 14/25x3y

15. 4/3a (a2 + b2 – 3c2)

Solution:

Let us simplify the given expression

4/3a (a2 + b2 – 3c2)

(4/3a × a2) + (4/3a × b2) – (4/3a × 3c2)

4/3a3 + 4/3ab2 – 4ac2

16. Find the product 24x2 (1-2x) and evaluate its value for x = 3

Solution:

Let us simplify the given expression

24x2 (1 – 2x)

(24x2× 1) – (24x2× 2x)

24x2 – 48x3

Now let us evaluate the expression when x = 3

24x2 – 48x3

24 × (3)2 – 48 × (3)3

24 × (9) – 48 × (27)

216 – 1296

-1080

17. Find the product -3y (xy+y2) and evaluate its value for x = 4 and y = 5

Solution:

Let us simplify the given expression

-3y (xy+y2)

(-3y × xy) + (-3y × y2)

-3xy2 – 3y3

Now let us evaluate the expression when x = 4 and y = 5

-3xy2 – 3y3

-3 × (4) × (5)2 – 3 × (5)3

-300 – 375

-675

18. Multiply -3/2x2y3 by (2x-y) and verify the answer for x = 1 and y = 2

Solution:

Let us simplify the given expression

-3/2x2y3 by (2x-y) 

(-3/2x2y3 × 2x) – (-3/2x2y3 × y)

-3x3y3 + 3/2x2y4

Now let us evaluate the expression when x = 1 and y = 2

-3x3y3 + 3/2x2y4

-3 × (1)4 × (2)3 + 3/2 × (1)2 × (2)4 

– 3 × (8) + 3 (8)

-24+24

0

19. Multiply the monomial by the binomial and find the value of each for x = -1, y = 0.25 and z = 0.005:
(i) 15y2 (2 – 3x)
(ii) -3x (y2 + z2)
(iii) z2 (x – y)
(iv) xz (x2 + y2)

Solution:

(i) 15y2 (2 – 3x)

Let us simplify the given expression

30y2 – 45xy2

By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000

30 × (25/100)2 – 45 × (-1) × (25/100)2

30 (1/16) + 45 (1/16)

15/8 + 45/16

(30+45)/16

75/16

(ii) -3x (y2 + z2)

Let us simplify the given expression

-3xy2 + -3xz2

By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000

-3× (-1) × (25/100)2 – 3 × (-1) × (5/1000)2

(3×25×25/100×100) + (3×5×5/1000×1000)

3/16 + 3/40000

39/200

(iii) z2 (x – y)

Let us simplify the given expression

z2x – z2y

By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000

z2 (x – y)

(5/1000)2 (-1 – 25/100)

(1/40000) (-100-25/100)

(1/40000) (-125/100)

(1/40000) (-5/4)

-5/160000

-1/32000

(iv) xz (x2 + y2)

Let us simplify the given expression

x3z + xzy2

By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000

x3z + xzy2

(-1)3 × (5/1000) + (-1) × (5/1000) × (25/100)2

-1/200 – 1/16 × 1/200

-1/200 – 1/3200

By taking LCM as 3200

(-16 -1)/3200

-17/3200

20. Simplify:

(i) 2x2 (x3 – x) – 3x (x4 + 2x) – 2 (x4 – 3x2)

(ii) x3y (x2 – 2x) + 2xy (x3 – x4)

(iii) 3a2 + 2 (a+2) – 3a (2a+1)

(iv) x (x+4) + 3x (2x2 -1) + 4x2 + 4

(v) a (b-c) – b (c-a) – c (a-b)

(vi) a (b-c) +b (c-a) + c (a-b)

(vii) 4ab (a-b) – 6a2 (b-b2) -3b2 (2a2 -a) + 2ab (b-a)

(viii) x2 (x2 + 1) – x3 (x + 1) – x (x3 – x)

(ix) 2a2 + 3a (1 – 2a3) + a (a + 1)

(x) a2 (2a – 1) + 3a + a3 – 8

(xi) 3/2x2 (x2 – 1) + 1/4x2 (x2 + x) – 3/4x (x3 – 1)

(xii) a2b (a-b2) + ab2(4ab – 2a2) – a3b(1-2b)

(xiii) a2b (a3– a + 1) – ab(a4 – 2a2 + 2a) – b(a3– a2 – 1)

Solution:

(i) 2x2 (x3 – x) – 3x (x4 + 2x) – 2 (x4 – 3x2)

Let us simplify the given expression

2x5 – 2x3 – 3x5 – 6x2 – 2x4 + 6x2

By grouping similar expressions we get,

2x5 – 3x5 – 2x3 – 2x4 – 6x2 + 6x2

-x5 – 2x4 – 2x3

(ii) x3y (x2 – 2x) + 2xy (x3 – x4)

Let us simplify the given expression

x5y – 2x4y + 2x4y – 2x5y

By grouping similar expressions we get,

-x5y – 2x5y

-x5y

(iii) 3a2 + 2 (a+2) – 3a (2a+1)

Let us simplify the given expression

3a2 + 2a + 4 – 6a2 – 3a

By grouping similar expressions we get,

3a2 – 6a2 + 2a – 3a + 4

-3a2 – a + 4

(iv) x (x+4) + 3x (2x2 -1) + 4x2 + 4

Let us simplify the given expression

x2 + 4x + 6x3 – 3x + 4x2 + 4

By grouping similar expressions we get,

6x3 + 5x2 + x + 4

(v) a (b-c) – b (c-a) – c (a-b)

Let us simplify the given expression

ab – ac – bc + ab – ca + bc

By grouping similar expressions we get,

2ab – 2ac

(vi) a (b-c) +b (c-a) + c (a-b)

Let us simplify the given expression

ab – ac + bc – ab + ac – bc

By grouping similar expressions we get,

0

(vii) 4ab (a-b) – 6a2 (b-b2) -3b2 (2a2 -a) + 2ab (b-a)

Let us simplify the given expression

4a2b – 4ab2 – 6a2b + 6a2b2 – 6a2b2 + 3ab2 + 2ab2 – 2a2b

By grouping similar expressions we get,

4a2b – 6a2b– 2a2b – 4ab2 + 3ab2 + 2ab2 + 6a2b2 – 6a2b2 

-4a2b + ab2 

(viii) x2 (x2 + 1) – x3 (x + 1) – x (x3 – x)

Let us simplify the given expression

x4 + x2 – x4 – x3 – x4 + x2

By grouping similar expressions we get,

x4 – x4 – x4 – x3 + x2 + x2

– x4 – x3 + 2x2 

(ix) 2a2 + 3a (1 – 2a3) + a (a + 1)

Let us simplify the given expression

2a2 + 3a – 6a4 + a2 + a

By grouping similar expressions we get,

-6a4 + 3a2 + 4a

(x) a2 (2a – 1) + 3a + a3 – 8

Let us simplify the given expression

2a3 – a2 + 3a + a3 – 8

By grouping similar expressions we get,

3a3 – a2 + 3a – 8

(xi) 3/2x2 (x2 – 1) + 1/4x2 (x2 + x) – 3/4x (x3 – 1)

Let us simplify the given expression

3/2x4 – 3/2x2 + 1/4x4 + 1/4x3 – 3/4x4 + 3/4x

By grouping similar expressions we get,

3/2x4 + 1/4x4 – 3/4x4 – 3/2x2 + 1/4x3 + 3/4x

4/4x4 + 1/4x3 – 3/2x2 + 3/4x

x4 + 1/4x3 – 3/2x2 + 3/4x

(xii) a2b (a-b2) + ab2(4ab – 2a2) – a3b(1-2b)

Let us simplify the given expression

a3b – a2b3 + 4a2b3 – 2a3b2 – a3b + 2a3b2

By grouping similar expressions we get,

-a2b3 + 4a2b3

3a2b3

(xiii) a2b (a3– a + 1) – ab(a4 – 2a2 + 2a) – b(a3– a2 – 1)

Let us simplify the given expression

a5b – a3b + a2b – a5b + 2a3b – 2a2b – ba3 + a2b + b

By grouping similar expressions we get,

a5b – a5b – a3b + 2a3b – ba3 + a2b – 2a2b + a2b + b

b


Access other Exercises of RD Sharma Solutions for Class 8 Maths Chapter 6 Algebraic Expressions and Identities

Exercise 6.1 Solutions

Exercise 6.2 Solutions

Exercise 6.3 Solutions

Exercise 6.5 Solutions

Exercise 6.6 Solutions

Exercise 6.7 Solutions

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