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

RD Sharma Solutions for Class 8 Mathematics

Created by: Abhishek Kapoor

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

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

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 ) Class 8 Notes | EduRev

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

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

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

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