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RD Sharma Class 8 Solutions Chapter 7 - Factorization (Ex 7.3) Exercise 7.3

Last updated date: 28th May 2024
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RD Sharma Class 8 Solutions Chapter 7 - Factorization (Ex 7.3) Exercise 7.3 - Free PDF

Free PDF download of RD Sharma Class 8 Solutions Chapter 7 - Factorization Exercise 7.3 solved by Expert Mathematics Teachers on All Chapter 7 - Factorization Ex 7.3 Questions with Solutions for RD Sharma Class 8 Maths to help you to revise the complete Syllabus and Score More marks. Register for online coaching for IIT JEE (Mains & Advanced) and other Engineering entrance exams. You can also register Online for Class 8 Science tuition on to score more marks in CBSE board examination. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students.

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Benefits of Downloading and Referring to the RD Sharma Class 8 Chapter 7 - Factorization

Written below are the benefits of downloading and referring to the RD Sharma class 8 chapter 7 - factorization. These are - 

  • These solutions are curated by the expert maths teacher with years of expertise in teaching and had taught many students in the past. By referring to these solutions students will be able to take part of their expertise.

  • Because of their experience in the field, teachers who are creating these solutions are well aware of the doubts that might arise. To tackle that extra notes are given at the end of the PDF.

  • Vedantu lets the students download the PDF of these solutions for absolutely free, Students can access them at any time they need.

FAQs on RD Sharma Class 8 Solutions Chapter 7 - Factorization (Ex 7.3) Exercise 7.3

1. Where can I access the solutions of other exercises in chapter 7 - Factorization?

Chapter 7 - factorization consists of 5 exercises in total, students can access the solutions to all the questions in those five exercises from the links given below to the relevant pages on the web platform of Vedantu. Given below are the links to the solutions of all four exercise questions of chapter 7 of RD Sharma of class 8 - 

Expert maths teachers, available at Vedantu have formulated these solutions with the utmost care to help students understand the concept better and secure good marks in their examination.

2. List all the important topics covered in Class 8 chapter 7 - Factorization?

Listed down here are some of the most important topics covered in the RD Sharma class 8 chapter 7 - Factorizations. - 

  • Definition of factors and factorization

  • Factors of monomials. The common and greatest common factor of monomials.

  • Factorization of algebraic expressions if a monomial is a common factor.

  • Factorization of algebraic expressions when a common Binomial factor occurs in each term.

  • Factorization by grouping the terms

  • factorization of quadratic polynomials.

  • factorization of quadratic polynomials with the method of perfect squares.

Students can learn more about these topics by solving questions from the RD Sharma book for class 8 chapter 7 exercises. 

3. Write down all the Identities that are used in chapter 7 - Factorization?

The basic identities that are used in the RD Sharma Class 8 chapter 7 - Factorization are written below:-

  1. (a + b)2 = a2 + 2ab + b2

  2. (a - b)2 = a2 - 2ab + b2

  3. (a + b)(a - b) = a2 b2

  4. (a + b + c)2 = a2 + b2  + c2 + 2ab + 2bc + 2ac

  5. (a + b)3 = a3+ 3a2b + 3a b2  + b3

  6. (a+x)(a+y) = a2 + (x+y)a + xy

Using all these identities, the factors of many different equations can be found.

4. Give examples of the questions that might be asked in the RD Sharma Class 8 Chapter 7 - Factorization (Ex 7 C) Exercise 7.3?

Example 1:- Factorize the following algebraic expressions.

1. (2x – 4y) (a + b) + (4x – 2y) (a + b)


(2x – 4y) (a + b) + (4x – 2y) (a + b) is Given


By taking (a + b) as common we get,


\[[(2x – 4y) + (4x – 2y)] (a + b)\]


\[[2x -4y + 4x – 2y] (a + b)\] 


\[[6x – 6y] (a + b)\]


\[(a + b) 6(x – y)\] Ans.


2.a (x – y) + 3b (y – x) + c (x – y)2


a (x – y) + 3b (y – x) + c (x – y)2      ….(Given)


By taking (-1) as common in the second term, we will get,


a (x – y) – 3b (x – y) + c (x – y)2


Then, by taking (x – y) as common we get,


\[[a – 3b + c(x – y)] (x – y)\]


\[(x – y) (a – 3b + cx – cy)\]   Ans.