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Class 8 RD Sharma Solutions – Chapter 6 Algebraic Expression and Identities – Exercise 6.4 | Set 2
  • Last Updated : 07 Apr, 2021

Chapter 6 Algebraic Expressions And Identities – Exercise 6.4 | Set 1

Question 11. Find the product of 1.5x (10x2y – 100xy2)

Solution:

Using Distributive law,

1.5x (10x2y – 100xy2) = (1.5x) × (10x2y) – (1.5x) × (100xy2)

= 15x2+1y – 150x1+1y2 = 15x3y – 150x2y2

Hence, the product is 15x3y – 150x2y2



Question 12. Find the product of 4.1xy (1.1x-y)

Solution:

Using Distributive law,

4.1xy (1.1x-y) = 4.1xy × 1.1x – 4.1xy × y

= 4.51x1+1y – 4.1xy1+1

= 4.51x2y – 4.1xy2

Hence, the product is 4.51x2y – 4.1xy2

Question 13. Find the product of 250.5xy (xz + y/10)

Solution:

Using Distributive law,



250.5xy (xz + y/10) = 250.5x1+1yz + 25.05xy1+1

= 250.5x2yz + 25.05xz2

Hence, the product is 250.5x2yz + 25.05xz2

Question 14. Find the product of 7x2y/5 (3xy2/5 + 2x/5)

Solution:

Using Distributive law,

7x2y/5 (3xy2/5 + 2x/5) = 21x2+1y1+2/25 + 14x2+1y/25

= 21x3y3/25 + 14x3y/25

Hence, the product is 21x3y3/25 + 14x3y/25

Question 15. Find the product of 4a/3 (a2 + b2 -3c2)

Solution:

Using Distributive law,

4a/3 (a2 + b2 -3c2) = 4a1+2/3 + 4ab2/3 – 4ac2

= 4a3/3 + 4ab2/3 – 4ac2

Hence, the product is 4a3/3 + 4ab2/3 – 4ac2

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

Solution:

Using Distributive law,

24x2 (1 – 2x) = 24x2 – 48x3

The product is 24x2 – 48x3

Now put x = 3

So, 24(3)2 – 48(3)3

= 216 – 1296 = -1080

The answer came out to be -1080

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

Solution:

Using Distributive law,

-3y (xy +y2) = -3xy2 – 3y3

The product is -3xy2 – 3y3

Now put x = 4, and y = 5

So, -3(4)(5)2 – 3(5)3

= -300 – 375 = -675

The answer came out to be -675

Question 18. Multiply – 3 x2y3/2 By (2x – y) and verify the answer for x = 1 and y = 2

Solution:

Using Distributive law,

– 3 x2y3/2(2x – y) = -3x3y3 + 3x2y4/2

The product is -3x3y3 + 3x2y4/2

Now put x = 1, and y = 2 and verifying the L.H.S and R.H.S

L.H.S = – 3 x2y3/2 (2x – y)

= -3(1)2(2)3/2 [2(1) – 2]

= 0

R.H.S = -3x3y3 + 3x2y4/2

= -3(13)(2)3 + 3(1)2(2)4/2 = -24 + 24 

= 0 

Since, L.H.S = R.H.S = 0

Hence verified

Question 19 Multiply the monomial by the binomial and find the value of each for x = -1, y = 0.25 and z =0.05 :
(i) 15y2 (2 – 3x)
(ii) -3x (y2 + z2)
(iii) z2 (x – y)
(iv) xz (x2 + y2)

Solution:

i) 15y2(2 – 3x)

Using Distributive law,

15y2 (2 – 3x) = 30y2 – 45xy2

The product of given monomial and binomial is 30y2 – 45xy2

Now put x = -1 and y = 0.25 

So, 30(0.25)2 – 45(-1)(0.25)2

= 1.8750 + 2.8125

= 4.6875

The answer will be 4.6875

ii) -3x(y2 + z2)

Using Distributive law,

-3x (y2 + z2) = -3xy2 – 3xz2

The product of given monomial and binomial is -3xy2 – 3xz2

Now put x = -1, y = 0.25 and z =0.05

-3(-1)(0.25)2 – 3(-1)(0.05)2

= 0.1875 + 0.0075 

= 0.1950

The answer will be 0.1950

iii) z2(x – y)

Using Distributive law,

z2(x – y) = z2x – z2y

The product of given monomial and binomial is z2x – z2y

Now put x = -1, y = 0.25 and z =0.05

= (0.05)2(-1) – (0.05)2(0.25)

= -0.0025 – 0.000625

= -0.003125

The answer will be -0.003125

iv) xz(x2 + y2)

Using Distributive law,

xz(x2 + y2) = x3z + xy2z

The product of given monomial and binomial is x3z + xy2z

Now put x = -1, y = 0.25 and z =0.05

= (-1)3(0.05) + (-1)(0.25)2(0.05)

= -0.05 – 0.003125

= -0.053125

The answer will be -0.053125

Question 20 Simplify :
(i) 2x2 (at1 – 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) a2b (a – b2) + ab2 (4ab – 2a2) – a3b (1 – 2b)
(xii)a2b (a3 – a + 1) – ab (a4 – 2a2 + 2a) – b (a3– a2 -1)

Solution:

(i) 2x2(at1 – x) – 3x(x4 + 2x) – 2(x4 – 3x2)

Using Distributive law,

2x2 (x3 – x) – 3x (x4 + 2x) -2 (x4 – 3x2) = 2x5 – 2x3 – 3x5 – 6x2 – 2x4 + 6x2

= -x5 – 2x4 – 2x3

Hence, the product is -x5 – 2x4 – 2x3

(ii) x3y (x2 – 2x) + 2xy (x3 – x4)

Using Distributive law,

x3y (x2 – 2x) + 2xy (x3 – x4) = x5y – 2x4y + 2x4y – 2x5y

= -x5

Hence, the product is -x5y

(iii) 3a2 + 2(a + 2) – 3a(2a + 1)

Using Distributive law,

3a2 + 2(a + 2) – 3a(2a + 1) = 3a2 + 2a + 4 – 6a2 – 3a

= – 3a2 – a + 4

Hence, the product is – 3a2 – a + 4

(iv) x(x + 4) + 3x(2x2 – 1) + 4x2 + 4

Using Distributive law,

x (x + 4) + 3x (2x2 – 1) + 4x2 + 4 = x2 + 4x + 6x3 -3x + 4x2 + 4 

= 6x3 + 5x2 + x + 4

Hence, the product is 6x3 + 5x2 + x + 4

(v) a(b – c) – b(c – a) – c(a – b)

Using Distributive law,

a (b – c) – b (c – a) – c (a – b) = ab – ac – bc + ab – ac + bc

= 2ab – 2ac

Hence, the product is 2ab – 2ac

(vi) a (b – c) + b (c – a) + c (a – b)

Using Distributive law,

a (b – c) + b (c – a) + c (a – b) = ab -ac +bc -ab +ac -bc = 0

Hence, the product is 0

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

Using Distributive law,

4ab (a – b) – 6a2 (b – b2) -3b2 (2a2 – a) + 2ab (b-a) = 4a2b – 4ab2 – 6a2b +6a2b2 -6a2b2+3ab2 +2ab2 -2a2b

= -4a2b + ab2

Hence, the product is -4a2b + ab2

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

Using Distributive law,

x2 (x2 + 1) – x3 (x + 1) – x (x3 – x) = x4 + x2 – x4 – x3 – x4 + x2 

= – x4 – x3+ 2x2

Hence, the product is – x4 – x3+ 2x2

(ix) 2a2 + 3a (1 – 2a3) + a (a + 1)

Using Distributive law,

2a2 + 3a (1 – 2a3) + a (a + 1) = 2a2 + 3a – 6a4 + a2+ a

= – 6a4 + 3a2 + 4a

Hence, the product is 6a4 + 3a2 + 4a

(x) a2 (2a – 1) + 3a + a3 – 8

Using Distributive law,

a2 (2a – 1) + 3a + a3 – 8 = 2a3 – a2 +3a + a3 – 8

= 3a3 – a2 + 3a -8

Hence, the product is 3a3 – a2 + 3a -8

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

Using Distributive law,

a2b (a – b2) + ab2 (4ab – 2a2) – a3b (1 – 2b) = a3b – a2b3 + 4a2b3 – 2a3b2 -a3b + 2a3b2

= 3a2b3

Hence, the product is 3a2b3

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

Using Distributive law,

a2b (a3 – a + 1) – ab (a4 – 2a2 + 2a) – b (a3– a2 -1) = a5b – a3b + a2b – a5b + 2a3b – 2a2b – a3b + a2b + b

= b

Hence, the product is b

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