RD Sharma Chapter 5 Class 9 Maths Exercise 5.1 Solutions is all about the Factorization of the Algebraic Expression. Students will get to know how to solve the various types of Algebraic Expressions like- monomials, binomials, trinomials, and polynomials using identities. Before going to any problems of this chapter, students are suggested to learn the formulas related to it, which helps them solve questions more quickly.
In the following RD Sharma Chapter 5 Class 9 Maths Exercise 5.1 Solutions PDF, learners understand the concept of the Factorization of the Algebraic Expressions through the solved problems based on the CBSE and RD Sharma of Class 9. The PDF is prepared by professionals with a complete explanation of each question.
Learn about RD Sharma Class 9 Chapter 5- Factorization of the Algebraic Expression
Download RD Sharma Chapter 5 Class 9 Maths Exercise 5.1 Solutions PDF
Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.1
Important Definitions RD Sharma Chapter 5 Class 9 Maths Exercise 5.1 Solutions
Go down to check the topics of Factorization of Algebraic Expressions from which simplification of the questions are given in the above PDF-
- Factorization by taking out the common factors
- Factorization by grouping the terms
- Factorization by making a perfect square
- Factorization by the difference of two squares
- Factorization of quadratic polynomials by splitting the middle term
Here are some examples with the solutions of the topics of Factorization of Algebraic Expressions-
Ques- a2 + 2ab + b2 – c2
Solution-
p2 + 2pq + q2 – r2
= (p2 + 2pq + q2) – r2
= (p + q)2 – (r) 2
Algebraic Identitty- p2 – q2 = (p + q) (p – q)
= (p + q + r) (p + q – r)
Therefore p2 + 2pq + q2 – r2 = (p + q + r) (p + q – r)
Ques- (a – b + c)2 + (b – c + a) 2 + 2(a – b + c) (b – c + a)
Solution-
(a – b + c)2 + (b – c + a) 2 + 2 (a – b + c) (b – c + a)
{Because p2 + q2 + 2pq = (p + q) 2}
Here p = a – b + c and q = b – c + a
= [a – b + c + b- c + a]2
= (2a)2
= 4a2
Ques- (x2 + 1/x2) – 4(x + 1/x) + 6
Solution:
(x2 + 1/x2) – 4(x + 1/x) + 6
= x2 + 1/x2 – 4x – 4/x + 4 + 2
= x2 + 1/x2 + 4 + 2 – 4/x – 4x
= (x2) + (1/x) 2 + (-2)2 + 2x(1/x) + 2(1/x)(-2) + 2(-2)x
As we know, x2 + y2 + z2 + 2xy + 2yz + 2zx = (x+y+z) 2
So, we can write;
= (x + 1/x + (-2 )) 2
or (x + 1/x – 2) 2
Therefore, (x2 + 1/x2) – 4(x + 1/x) + 6 = (x + 1/x – 2) 2
Ques- 2a2 + 2√6 ab + 3b2
Solution:
2a2 + 2√6 ab + 3b2
Above expression can be written as ( √2a )2 + 2 × √2a × √3b + ( √3b)2
As we know, ( p + q ) 2 = p2 + q2 + 2pq
Here p = √2a and q = √3b
= (√2a + √3b )2
Therefore, 2a2 + 2√6 ab + 3b2 = (√2a + √3b )2