Class 8 Exam  >  Class 8 Notes  >  RD Sharma Solutions for Class 8 Mathematics  >  RD Sharma Solutions for Class 8 Math Chapter 7 - Factorization (Part-3)

RD Sharma Solutions for Class 8 Math Chapter 7 - Factorization (Part-3) | RD Sharma Solutions for Class 8 Mathematics PDF Download

Question 1: Factorize each of the following algebraic expressions:
6x(2xy) + 7y(2xy)
Answer 1: 6x(2xy)+7y(2xy) 

= (6x+7y)(2xy)         [Taking (2xy) as the common factor] 
Question 2:Factorize each of the following algebraic expressions:
2r(y − x) + s(x − y) 
Answer 2: 
2r(yx)+s(xy)
=2r(yx)s(yx)      [(xy)=(yx)] 
=(2rs)(yx)              [Taking (yx) as the common factor] 

Question 3: Factorize each of the following algebraic expressions:
7a(2x − 3) + 3b(2x − 3) 
Answer 3:
7a(2x3)+3b(2x3) 
=(7a+3b)(2x3)            [Taking (2x3) as the common factor]
Question 4: Factorize each of the following algebraic expressions:
9a(6a − 5b) −12a2(6a − 5b)
Answer 4:
9a(6a5b)12a2(6a5b) 
=(9a12a2)(6a5b)     [Taking (6a5b) as the common factor] 
=3a(34a)(6a5b)       [Taking 3a as the common factor of the quadratic (9a12a2)

Question 5: Factorize each of the following algebraic expressions:
5(x − 2y)2 + 3(x − 2y) 
Answer 5:
5(x2y)2+3(x2y)
=[5(x2y)+3](x2y)     [Taking (x2y) as the common factor] (5x10y+3)(x2y)

Question 6: Factorize each of the following algebraic expressions: 
16(2l − 3m)2 −12(3m − 2l) 
Answer 6: 16(2l3m)212(3m2l) 

=16(2l3m)2+12(2l3m)            [(3m2l)=(2l3m)] 
= [16(2l3m)+12](2l3m)           [Taking (2l3m) as the common factor] 
=4[4(2l3m)+3](2l3m)              {Taking 4 as the common factor of [16(2l3m)+12]} 
=4(8l12m+3)(2l3m) 
Question 7: Factorize each of the following algebraic expressions: 
3a(x − 2y) −b(x − 2y) 
Answer 7: 3a(x2y)b(x2y) 

=(3ab)(x2y)           [Taking (x2y) as the common factor] 

Question 8: Factorize each of the following algebraic expressions: 
a2(x + y) +b2(x + y) +c2(x + y) 
Answer 8: a2(x+y)+b2(x+y)+c2(x+y) 

=(a2+b2+c2)(x+y)                      [Taking (x+y) as the common factor]
Question 9: Factorize each of the following algebraic expressions: 
(x − y)2 + (x − y) 
Answer 9: (xy)2+(xy) 

=(xy)(xy)+(xy)   [Taking (xy) as the common factor] 
= (xy+1)(xy) 
Question 10: Factorize each of the following algebraic expressions: 
6(a + 2b) −4(a + 2b)2 

Answer 10: 6(a+2b)4(a+2b)2 

=[64(a+2b)](a+2b)      [Taking (a+2b) as the common factor] 
= 2[32(a+2b)](a+2b)   {Taking 2 as the common factor of [64(a+2b)]} 
=2(32a4b)(a+2b) 
Question 11: Factorize each of the following algebraic expressions:
a(x − y) + 2b(y − x) + c(x − y)2 
Answer 11: a(xy)+2b(yx)+c(xy)2 

=a(xy)2b(xy)+c(xy)2   [(yx)=(xy)] 
= [a2b+c(xy)](xy) 
= (a2b+cxcy)(xy) 
Question 12: Factorize each of the following algebraic expressions:
−4(x − 2y)2 + 8(x −2y) 
Answer 12: 
4(x2y)2+8(x2y) 
= [4(x2y)+8](x2y)   [Taking (x2y) as the common factor] 
= 4[(x2y)+2](x2y)   {Taking 4 as the common factor of [4(x2y)+8]} 
= 4(2yx+2)(x2y) 
Question 13: Factorize each of the following algebraic expressions:
x3(a − 2b) + x2(a − 2b)
Answer 13: x3(a2b)+x2(a2b) 

=(x3+x2)(a2b)       [Taking (a2b) as the common factor] 
= x2(x+1)(a2b)      [Taking x2 as the common factor of (x3+x2)]
Question 14: Factorize each of the following algebraic expressions:

(2x − 3y)(a + b) + (3x − 2y)(a + b) 
Answer 14: (2x3y)(a+b)+(3x2y)(a+b) 

=(2x3y+3x2y)(a+b)   [Taking (a+b) as the common factor] 
=(5x5y)(a+b) 
=5(xy)(a+b)                      [Taking 5 as the common factor of (5x5y)]
Question 15: Factorize each of the following algebraic expressions: 
4(x + y) (3a − b) +6(x + y) (2b − 3a) 
Answer 15: 4(x+y)(3ab)+6(x+y)(2b3a)

=2(x+y)[2(3ab)+3(2b3a)]                {Taking [2 (x+y)] as the common factor} 
=2(x+y)(6a2b+6b9a) 
=2(x+y)(4b3a)

The document RD Sharma Solutions for Class 8 Math Chapter 7 - Factorization (Part-3) | RD Sharma Solutions for Class 8 Mathematics is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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FAQs on RD Sharma Solutions for Class 8 Math Chapter 7 - Factorization (Part-3) - RD Sharma Solutions for Class 8 Mathematics

1. What is factorization?
Ans. Factorization is the process of breaking down a number or an algebraic expression into its factors. Factors are the numbers or expressions that multiply together to give the original number or expression.
2. How do you factorize a quadratic expression?
Ans. To factorize a quadratic expression, we need to find two binomials whose product is equal to the given expression. We can use various methods such as grouping, trial and error, or using the quadratic formula to find the factors of the expression.
3. Can all numbers be factorized?
Ans. Yes, all numbers can be factorized. Every number can be expressed as a product of its prime factors. This process is known as prime factorization. For example, the number 24 can be factorized as 2 x 2 x 2 x 3.
4. What is the difference between prime factorization and factorization?
Ans. Prime factorization is the process of expressing a number as a product of its prime factors, while factorization refers to breaking down a number or an expression into its factors, which may or may not be prime.
5. How do you factorize a polynomial expression?
Ans. To factorize a polynomial expression, we use various techniques such as finding common factors, using the factor theorem, or using the quadratic formula. The goal is to break down the expression into its simplest form by finding its factors.
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