# Class 8 RD Sharma Solutions – Chapter 6 Algebraic Expression and Identities – Exercise 6.4 | Set 2

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

**Question 11. Find the product of 1.5x (10x**^{2}y â€“ 100xy^{2})

^{2}y â€“ 100xy

^{2})

**Solution:**

Using Distributive law,

1.5x (10x

^{2}y â€“ 100xy^{2}) = (1.5x) Ã— (10x^{2}y) – (1.5x) Ã— (100xy^{2})= 15x

^{2+1}y – 150x^{1+1}y^{2}= 15x^{3}y – 150x^{2}y^{2}

Hence, the product is 15x^{3}y – 150x^{2}y^{2}

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

**Solution:**

Using Distributive law,

4.1xy (1.1x-y) = 4.1xy Ã— 1.1x – 4.1xy Ã— y

= 4.51x

^{1+1}y – 4.1xy^{1+1}= 4.51x

^{2}y – 4.1xy^{2}

Hence, the product is 4.51x^{2}y – 4.1xy^{2}

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

**Solution:**

Using Distributive law,

250.5xy (xz + y/10) = 250.5x

^{1+1}yz + 25.05xy^{1+1}= 250.5x

^{2}yz + 25.05xz^{2}

Hence, the product is 250.5x^{2}yz + 25.05xz^{2}

**Question 14. Find the product of 7x**^{2}y/5 (3xy^{2}/5 + 2x/5)

^{2}y/5 (3xy

^{2}/5 + 2x/5)

**Solution:**

Using Distributive law,

7x

^{2}y/5 (3xy^{2}/5 + 2x/5) = 21x^{2+1}y^{1+2}/25 + 14x^{2+1}y/25= 21x

^{3}y^{3}/25 + 14x^{3}y/25

Hence, the product is 21x^{3}y^{3}/25 + 14x^{3}y/25

**Question 15. Find the product of 4a/3 (a**^{2} + b^{2} -3c^{2})

^{2}+ b

^{2}-3c

^{2})

**Solution:**

Using Distributive law,

4a/3 (a

^{2}+ b^{2}-3c^{2}) = 4a^{1+2}/3 + 4ab^{2}/3 – 4ac^{2}= 4a

^{3}/3 + 4ab^{2}/3 – 4ac^{2}

Hence, the product is 4a^{3}/3 + 4ab^{2}/3 – 4ac^{2}

**Question 16. Find the product 24x**^{2} (1 â€“ 2x) and evaluate its value for x = 3.

^{2}(1 â€“ 2x) and evaluate its value for x = 3.

**Solution:**

Using Distributive law,

24x

^{2}(1 – 2x) = 24x^{2}– 48x^{3}

The product is 24x^{2}– 48x^{3}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 +y**^{2}) and find its value for x = 4, and y = 5.

^{2}) and find its value for x = 4, and y = 5.

**Solution:**

Using Distributive law,

-3y (xy +y

^{2}) = -3xy^{2}– 3y^{3}

The product is -3xy^{2}– 3y^{3}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 â€“ 3 x**^{2}y^{3}/2 By (2x – y) and verify the answer for x = 1 and y = 2

^{2}y

^{3}/2 By (2x – y) and verify the answer for x = 1 and y = 2

**Solution:**

Using Distributive law,

â€“ 3 x

^{2}y^{3}/2(2x – y) = -3x^{3}y^{3}+ 3x^{2}y^{4}/2

The product is -3x^{3}y^{3}+ 3x^{2}y^{4}/2Now put x = 1, and y = 2 and verifying the L.H.S and R.H.S

L.H.S = â€“ 3 x

^{2}y^{3}/2 (2x – y)= -3(1)

^{2}(2)^{3}/2 [2(1) – 2]= 0

R.H.S = -3x

^{3}y^{3}+ 3x^{2}y^{4}/2= -3(1

^{3})(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) 15y**^{2} (2 â€“ 3x)

**(ii) -3x (y**^{2} + z^{2})

**(iii) z**^{2} (x â€“ y)

**(iv) xz (x**^{2} + y^{2})

^{2}(2 â€“ 3x)

^{2}+ z

^{2})

^{2}(x â€“ y)

^{2}+ y

^{2})

**Solution:**

**i) 15y ^{2}(2 â€“ 3x)**

Using Distributive law,

15y

^{2}(2 â€“ 3x) = 30y^{2}– 45xy^{2}

The product of given monomial and binomial is 30y^{2}– 45xy^{2}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(y ^{2} + z^{2})**

Using Distributive law,

-3x (y

^{2}+ z^{2}) = -3xy^{2}– 3xz^{2}

The product of given monomial and binomial is -3xy^{2}– 3xz^{2}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) z ^{2}(x â€“ y)**

Using Distributive law,

z

^{2}(x â€“ y) = z^{2}x – z^{2}y

The product of given monomial and binomial is z^{2}x – z^{2}yNow 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(x ^{2} + y^{2})**

Using Distributive law,

xz(x

^{2}+ y^{2}) = x^{3}z + xy^{2}z

The product of given monomial and binomial is x^{3}z + xy^{2}zNow 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) 2x**^{2} (at^{1} â€“ x) â€“ 3x (x^{4} + 2x) -2 (x^{4} â€“ 3x^{2})

**(ii) x**^{3}y (x^{2} â€“ 2x) + 2xy (x^{3} â€“ x^{4})

**(iii) 3a**^{2} + 2 (a + 2) â€“ 3a (2a + 1)

**(iv) x (x + 4) + 3x (2x**^{2} â€“ 1) + 4x^{2} + 4

**(v) a (b-c) â€“ b (c â€“ a) â€“ c (a â€“ b)**

**(vi) a (b â€“ c) + b (c â€“ a) + c (a â€“ b)**

**(vii) 4ab (a â€“ b) â€“ 6a**^{2} (b â€“ b^{2}) -3b^{2} (2a^{2} â€“ a) + 2ab (b-a)

**(viii) x**^{2} (x^{2} + 1) â€“ x^{3} (x + 1) â€“ x (x^{3} â€“ x)

**(ix) 2a**^{2} + 3a (1 â€“ 2a^{3}) + a (a + 1)

**(x) a**^{2} (2a â€“ 1) + 3a + a^{3} â€“ 8

**(xi) a**^{2}b (a â€“ b^{2}) + ab^{2} (4ab â€“ 2a^{2}) â€“ a^{3}b (1 â€“ 2b)

**(xii)a**^{2}b (a^{3} â€“ a + 1) â€“ ab (a^{4} â€“ 2a^{2} + 2a) â€“ b (a^{3}â€“ a^{2} -1)

^{2}(at

^{1}â€“ x) â€“ 3x (x

^{4}+ 2x) -2 (x

^{4}â€“ 3x

^{2})

^{3}y (x

^{2}â€“ 2x) + 2xy (x

^{3}â€“ x

^{4})

^{2}+ 2 (a + 2) â€“ 3a (2a + 1)

^{2}â€“ 1) + 4x

^{2}+ 4

^{2}(b â€“ b

^{2}) -3b

^{2}(2a

^{2}â€“ a) + 2ab (b-a)

^{2}(x

^{2}+ 1) â€“ x

^{3}(x + 1) â€“ x (x

^{3}â€“ x)

^{2}+ 3a (1 â€“ 2a

^{3}) + a (a + 1)

^{2}(2a â€“ 1) + 3a + a

^{3}â€“ 8

^{2}b (a â€“ b

^{2}) + ab

^{2}(4ab â€“ 2a

^{2}) â€“ a

^{3}b (1 â€“ 2b)

^{2}b (a

^{3}â€“ a + 1) â€“ ab (a

^{4}â€“ 2a

^{2}+ 2a) â€“ b (a

^{3}â€“ a

^{2}-1)

**Solution:**

**(i) 2x ^{2}(at^{1} â€“ x) â€“ 3x(x^{4} + 2x) – 2(x^{4} â€“ 3x^{2})**

Using Distributive law,

2x

^{2}(x^{3}– x) – 3x (x^{4}+ 2x) -2 (x^{4}– 3x^{2}) = 2x^{5}– 2x^{3}– 3x^{5}– 6x^{2}– 2x^{4}+ 6x^{2}= -x

^{5}– 2x^{4}– 2x^{3}

Hence, the product is -x^{5}– 2x^{4}– 2x^{3}

**(ii) x ^{3}y (x^{2} – 2x) + 2xy (x^{3} – x^{4})**

Using Distributive law,

x

^{3}y (x^{2}â€“ 2x) + 2xy (x^{3}â€“ x^{4}) = x^{5}y – 2x^{4}y + 2x^{4}y – 2x^{5}y= -x

^{5}y

Hence, the product is -x^{5}y

**(iii) 3a ^{2} + 2(a + 2) – 3a(2a + 1)**

Using Distributive law,

3a

^{2}+ 2(a + 2) – 3a(2a + 1) = 3a^{2 }+ 2a + 4 – 6a^{2}– 3a= – 3a

^{2 }– a + 4

Hence, the product is – 3a^{2 }– a + 4

**(iv) x(x + 4) + 3x(2x ^{2} â€“ 1) + 4x^{2} + 4**

Using Distributive law,

x (x + 4) + 3x (2x

^{2}â€“ 1) + 4x^{2}+ 4 = x^{2}+ 4x + 6x^{3}-3x + 4x^{2}+ 4= 6x

^{3}+ 5x^{2}+ x + 4

Hence, the product is 6x^{3}+ 5x^{2}+ x + 4

**(v) a(b – c) – b(c – a) – c(a – b)**

Using Distributive law,

a (b – c) – b (c â€“ a) â€“ c (a â€“ b) = ab – ac – bc + ab – ac + bc

= 2ab – 2ac

Hence, the product is 2ab – 2ac

**(vi) a (b â€“ c) + b (c â€“ a) + c (a â€“ b)**

Using Distributive law,

a (b â€“ c) + b (c â€“ a) + c (a â€“ b) = ab -ac +bc -ab +ac -bc = 0

Hence, the product is 0

**(vii) 4ab(a – b) – 6a ^{2}(b – b^{2}) – 3b^{2}(2a^{2} – a) + 2ab(b – a)**

Using Distributive law,

4ab (a â€“ b) â€“ 6a

^{2}(b â€“ b^{2}) -3b^{2}(2a^{2}â€“ a) + 2ab (b-a) = 4a^{2}b – 4ab^{2}– 6a^{2}b +6a^{2}b^{2}-6a^{2}b^{2}+3ab^{2}+2ab^{2}-2a^{2}b= -4a

^{2}b + ab^{2}

Hence, the product is -4a^{2}b + ab^{2}

**(viii) x ^{2} (x^{2} + 1) â€“ x^{3} (x + 1) â€“ x (x^{3} â€“ x)**

Using Distributive law,

x

^{2}(x^{2}+ 1) â€“ x^{3}(x + 1) â€“ x (x^{3}â€“ x) = x^{4}+ x^{2}– x^{4}– x^{3}– x^{4}+ x^{2}= – x

^{4 }– x^{3}+ 2x^{2}

Hence, the product is – x^{4 }– x^{3}+ 2x^{2}

**(ix) 2a ^{2} + 3a (1 â€“ 2a^{3}) + a (a + 1)**

Using Distributive law,

2a

^{2}+ 3a (1 â€“ 2a^{3}) + a (a + 1) = 2a^{2}+ 3a – 6a^{4}+ a^{2}+ a= – 6a

^{4}+ 3a^{2}+ 4a

Hence, the product is–6a^{4}+ 3a^{2}+ 4a

**(x) a ^{2} (2a â€“ 1) + 3a + a^{3} â€“ 8**

Using Distributive law,

a

^{2}(2a â€“ 1) + 3a + a^{3}â€“ 8 = 2a^{3}– a^{2}+3a + a^{3}â€“ 8= 3a

^{3}– a^{2}+ 3a -8

Hence, the product is 3a^{3}– a^{2}+ 3a -8

**(xi) a ^{2}b (a â€“ b^{2}) + ab^{2} (4ab â€“ 2a^{2}) â€“ a^{3}b (1 â€“ 2b)**

Using Distributive law,

a

^{2}b (a â€“ b^{2}) + ab^{2}(4ab â€“ 2a^{2}) â€“ a^{3}b (1 â€“ 2b) = a^{3}b – a^{2}b^{3}+ 4a^{2}b^{3}– 2a^{3}b^{2}-a^{3}b + 2a^{3}b^{2}= 3a

^{2}b^{3}

Hence, the product is3a^{2}b^{3}

**(xii) a ^{2}b (a^{3} â€“ a + 1) â€“ ab (a^{4} â€“ 2a^{2} + 2a) â€“ b (a^{3}â€“ a^{2} -1)**

Using Distributive law,

a

^{2}b (a^{3}â€“ a + 1) â€“ ab (a^{4}â€“ 2a^{2}+ 2a) â€“ b (a^{3}â€“ a^{2}-1) = a^{5}b – a^{3}b + a^{2}b – a^{5}b + 2a^{3}b – 2a^{2}b – a^{3}b + a^{2}b + b= b

Hence, the product isb

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