Students can avail of the free PDF of RD Sharma Solutions for Class 8 Maths Exercise 6.4 of Chapter 6 Algebraic Expressions and Identities which is provided here. In Exercise 6.4 of Chapter 6, Algebraic Expressions and Identities, we will discuss problems based on the multiplication of a monomial and a binomial. The 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 will help students improve their speed in solving the questions quickly in exams. The PDF of RD Sharma Solutions for Class 8 Chapter 6 Exercise 6.4 can be downloaded easily from the links provided below.
RD Sharma Solutions for Class 8 Maths Exercise 6.4 Chapter 6 Algebraic Expressions and Identities
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 6.4 Chapter 6 Algebraic Expressions and Identities
EXERCISE 6.4 PAGE NO: 6.21
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
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