Class 8 RD Sharma Solutions – Chapter 6 Algebraic Expressions And Identities – Exercise 6.4 | Set 1

Question 1. Find the product 2a³(3a + 5b)

Solution:

Using Distributive law,

2a³ (3a + 5b) = 2a³ × 3a + 2a³ × 5b

= 6a³⁺¹ + 10a³b = 6a⁴ + 10a³b

Hence, the product is 6a⁴ + 10a³b

Question 2. Find the product -11a(3a + 2b)

Solution:

Using Distributive law,

-11a (3a + 2b) = -11a × 3a + (-11a) × 2b

= -33a¹⁺¹ – 22ab = -33a² – 22ab

Hence, the product is -33a² – 22ab

Question 3. Find the product -5a(7a – 2b)

Solution:

Using Distributive law,

-5a (7a – 2b) = -5a × 7a – (-5a) × 2b

= -35a¹⁺¹ + 10ab = -35a² + 10ab

Hence, the product is -35a² + 10ab

Question 4. Find the product -11y²(3y + 7)

Solution:

Using Distributive law,

-11y² (3y + 7) = -11y² × 3y + (-11y²) × 7

= -33y²⁺¹ – 77y² = -33y³ – 77y²

Hence, the product is -33y³ – 77y²

Question 5. Find the product of 6x/5(x³+y³)

Solution:

Using Distributive law,

6x/5 (x³+y³) = 6x/5 × x³ + 6x/5 × y³

= 6x³⁺¹/5 + 6xy³/5 = 6x⁴/5 + 6xy³/5

Hence, the product is 6x⁴/5 + 6xy³/5

Question 6. Find the product of xy(x³ – y³)

Solution:

Using Distributive law,

xy (x³-y³) = xy × x³ – xy × y³

= x³⁺¹y – xy³⁺¹ = x⁴y – xy⁴

Hence, the product is x⁴y – xy⁴

Question 7. Find the product of 0.1y (0.1x⁵ + 0.1y)

Solution:

Using Distributive law,

0.1y (0.1x⁵ + 0.1y) = 0.1y × 0.1x⁵ + 0.1y × 0.1y 

= 0.01x⁵y + 0.01y²

Hence, the product is 0.01x⁵y + 0.01y²

Question 8. Find the product of (-7ab²c/4 – 6a²c²/25) (-50a²b²c²)

Solution:

Using Distributive law,

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

= 175a²⁺¹b²⁺²c²⁺¹/2 + 12a²⁺²b²c²⁺²

= 175a³b⁴c³/2 + 12a⁴b²c⁴

Hence, the product is 175a³b⁴c³/2 + 12a⁴b²c⁴

Question 9. Find the product of -8xyz/27 (3xyz²/2 – 9xy²z³/4)

Solution:

Using Distributive law,

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

= -4x¹⁺¹y¹⁺¹z¹⁺²/9 + 2x¹⁺¹y¹⁺²z¹⁺³/3
= -4x²y²z³/9 + 2x²y³z⁴/3

Hence, the product is -4x²y²z³/9 + 2x²y³z⁴/3

Question 10. Find the product of (-4xyz/27) [9x²yz/2 – 3xyz²/4]

Solution:

Using Distributive law,

(-4xyz/27) [9x²yz/2 – 3xyz²/4] = (-4xyz/27) × (9x²yz/2) + (-4xyz/27) × (3xyz²/4)

= -2x¹⁺²y¹⁺¹z¹⁺¹/3 + 1x¹⁺¹y¹⁺¹z¹⁺²/9 

= -2x³y²z²/3 + 1x²y²z³/9 

Hence, the product is –2x³y²z²/3 + 1x²y²z³/9 

 Chapter 6 Algebraic Expression and Identities – Exercise 6.4 | Set 2

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