RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Class 8: RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

The document RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8

Question 1: Multiply:
(5x + 3) by (7x + 2)

Answer 1: To multiply, we will use distributive law as follows:
(5x+3)(7x+2)=5x(7x+2)+3(7x+2)=(5x×7x+5x×2)+(3×7x+3×2)=(35x2+10x)+(21x+6)=35x2+10x+21x+6=35x2+31x+6
Thus, the answer is 35x2+31x+635x2+31x+6. 

Question 2: Multiply:
(2x + 8) by (x − 3)

Answer 2: To multiply the expressions, we will use the distributive law in the following way:
(2x+8)(x3)=2x(x3)+8(x3)=(2x×x2x×3)+(8x8×3)=(2x26x)+(8x24)=2x26x+8x24=2x2+2x242x+8x-3=2xx-3+8x-3=2x×x-2x×3+8x-8×3=2x2-6x+8x-24=2x2-6x+8x-24=2x2+2x-2Thus, the answer is 2x2+2x242x2+2x-24. 

Question 3: Multiply:
(7x + y) by (x + 5y)

Answer 3: To multiply, we will use distributive law as follows:
(7x+y)(x+5y)=7x(x+5y)+y(x+5y)=7x2+35xy+xy+5y2=7x2+36xy+5y27x+yx+5y=7xx+5y+yx+5y=7x2+35xy+xy+5y2=7x2+36xy+5y2
Thus, the answer is 7x2+36xy+5y27x2+36xy+5y2. 

Question 4: Multiply:
(a − 1) by (0.1a2 + 3)

Answer 4: To multiply, we will use distributive law as follows:
(a1)(0.1a2+3)=0.1a2(a1)+3(a1)=0.1a30.1a2+3a3a-10.1a2+3=0.1a2a-1+3a-1=0.1a3-0.1a2+3a-3
Thus, the answer is 0.1a30.1a2+3a30.1a3-0.1a2+3a-3. 

Question 5: Multiply:
(3x2 + y2) by (2x2 + 3y2)

Answer 5: To multiply, we will use distributive law as follows:
(3x2+y2)(2x2+3y2)=3x2(2x2+3y2)+y2(2x2+3y2)=6x4+9x2y2+2x2y2+3y4=6x4+11x2y2+3y43x2+y22x2+3y2=3x22x2+3y2+y22x2+3y2=6x4+9x2y2+2x2y2+3y4=6x4+11x2y2+3yThus, the answer is 6x4+11x2y2+3y46x4+11x2y2+3y4. 

Question 6: Multiply:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Answer 6: To multiply, we will use distributive law as follows: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Thus, the answer is RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Question 7: Multiply:
(x6y6) by (x2 + y2)

Answer 7: To multiply, we will use distributive law as follows: 

(x6y6)(x2+y2)=x6(x2+y2)y6(x2+y2)=(x8+x6y2)(y6x2+y8)=x8+x6y2y6x2y8 Thus, the answer is x8+x6y2y6x2y8x8+x6y2-y6x2-y8. 

Question 8: Multiply:
(x2 + y2) by (3a + 2b)

Answer 8: To multiply, we will use distributive law as follows: 

(x2+y2)(3a+2b)=x2(3a+2b)+y2(3a+2b)=3ax2+2bx2+3ay2+2by2 Thus, the answer is 3ax2+2bx2+3ay2+2by23ax2+2bx2+3ay2+2by2. 

Question 9: Multiply:
[−3d + (−7f)] by (5d + f)

Answer 9: To multiply, we will use distributive law as follows:
[3d+(7f)](5d+f)=(3d)(5d+f)+(7f)(5d+f)=(15d23df)+(35df7f2)=15d23df35df7f2=15d238df7f2
Thus, the answer is 15d238df7f2-15d2-38df-7f2. 

Question 10: Multiply:
(0.8a − 0.5b) by (1.5a − 3b)

Answer 10: To multiply, we will use distributive law as follows:
(0.8a0.5b)(1.5a3b)=0.8a(1.5a3b)0.5b(1.5a3b)=1.2a22.4ab0.75ab+1.5b2=1.2a23.15ab+1.5b2
Thus, the answer is 1.2a23.15ab+1.5b21.2a2-3.15ab+1.5b2. 

Question 11: Multiply:
(2x2y2 − 5xy2) by (x2y2)

Answer 11: To multiply, we will use distributive law as follows:

(2x2y25xy2)(x2y2)=2x2y2(x2y2)5xy2(x2y2)=2x4y22x2y45x3y2+5xy4 Thus, the answer is 2x4y22x2y45x3y2+5xy42x4y2-2x2y4-5x3y2+5xy4. 

Question 12: Multiply:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Answer 12: To multiply the expressions, we will use the distributive law in the following way: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Thus, the answer is RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Question 13: Multiply:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Answer 13: To multiply, we will use distributive law as follows: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Thus, the answer is RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Question 14: Multiply:

(3x2y − 5xy2) by RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Answer 14: To multiply, we will use distributive law as follows: 

(3x2y5xy2) (1/5 x2+1/3 y2) 

1/5 x2(3x2y5xy2)+1/3 y2(3x2y5xy2) 

= 3/5 x4yx3y2+x2y35/3 xy4 

Thus, the answer is 3/5 x4yx3y2+x2y35/3 xy4 

Question 15: Multiply:
(2x2 − 1) by (4x3 + 5x2)

Answer 15: To multiply, we will use distributive law as follows:
(2x21)(4x3+5x2)=2x2(4x3+5x2)1(4x3+5x2)=8x5+10x44x35x2
Thus, the answer is 8x5+10x44x35x28x5+10x4-4x3-5x2. 

Question 16: (2xy + 3y2) (3y2 − 2)

Answer 16: To multiply, we will use distributive law as follows:
(2xy+3y2)(3y22)=2xy(3y22)+3y2(3y22)=6xy34xy+9y46y2=9y4+6xy36y24xy
Thus, the answer is 9y4+6xy36y24xy9y4+6xy3-6y2-4xy. 

Question 17: Find the following product and verify the result for x = − 1, y = − 2:
(3x − 5y) (x + y)

Answer 17: To multiply, we will use distributive law as follows:
(3x5y)(x+y)=3x(x+y)5y(x+y)=3x2+3xy5xy5y2=3x22xy5y2
 (3x5y)(x+y)=3x22xy5y23x-5yx+y=3x2-2xy-5y2. 

Now, we put x = -1 and y = -2 on both sides to verify the result.

LHS=(3x5y)(x+y)={3(1)5(2)}{1+(2)}=(3+10)(3)=(7)(3)=21 RHS=3x22xy5y2=3(1)22(1)(2)5(2)2=3×145×4=3420=21 Because LHS is equal to RHS, the result is verified.
Thus, the answer is 3x22xy5y23x2-2xy-5y2. 
 

Question 18: Find the following product and verify the result for x = − 1, y = − 2:
(x2y 1) (3 2x2y)

Answer 18: To multiply, we will  use distributive law as follows: 

(x2y1)(32x2y)=x2y(32x2y)1×(32x2y)=3x2y2x4y23+2x2y=5x2y2x4y23  (x2y1)(32x2y)=5x2y2x4y23x2y-13-2x2y=5x2y-2x4y2-3

Now, we put x = -1 and y = -2 on both sides to verify the result. 

LHS = (x2y1)(32x2y)=[(1)2(2)1][32(1)2(2)]=[1×(2)1][32×1×(2)]=(21)(3+4)=3×7=21RHS=5x2y2x4y23=5(1)2(2)2(1)4(2)23=[5×1×(2)][2×1×4]3=1083=21  Because LHS is equal to RHS, the result is verified.
Thus, the answer is 5x2y2x4y235x2y-2x4y2-3. 

Question 19: Find the following product and verify the result for x = − 1, y = − 2:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Answer 19: To multiply, we will use distributive law as follows:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Now, we will put x = -1 and y = -2 on both the sides to verify the result. 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

=- 119/225

RHS= RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

= - 119/225

Because LHS is equal to RHS, the result is verified. 

Thus, the answer is RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Question 20: Simplify:
x2(x + 2y) (x − 3y)

Answer 20: To simplify, we will proceed as follows: 

x2(x+2y)(x3y)=[x2(x+2y)](x3y)=(x3+2x2y)(x3y)=x3(x3y)+2x2y(x3y)=x43x3y+2x3y6x2y2=x4x3y6x2y2 Thus, the answer is x4x3y6x2y2x4-x3y-6x2y2. 

Question 21: Simplify:
(x2 − 2y2) (x + 4y) x2y2

Answer 21: To simplify, we will proceed as follows: 

(x22y2)(x+4y)x2y2=[x2(x+4y)2y2(x+4y)]x2y2=(x3+4x2y2xy28y3)x2y2=x5y2+4x4y32x3y48x2y5 Thus, the answer is x5y2+4x4y32x3y48x2y5x5y2+4x4y3-2x3y4-8x2y5. 

Question 22: Simplify:
a2b2(a + 2b)(3a + b)

Answer 22: To simplify, we will proceed as follows: 

a2b2(a+2b)(3a+b)=[a2b2(a+2b)](3a+b)=(a3b2+2a2b3)(3a+b)=3a(a3b2+2a2b3)+b(a3b2+2a2b3)=3a4b2+6a3b3+a