# RD Sharma Solutions Class 9 Maths Chapter 5 – Factorization Of Algebraic Expressions (Updated for 2021-22) RD Sharma Solutions Class 9 Maths Chapter 5 – Factorization Of Algebraic Expressions: Looking for some quality study material other than NCERT guides. Well, this is the right place you are looking for. We are here to present you RD Sharma Solutions Class 9 Maths Chapter 5. This is certainly one of the best choices to finish chapter 5 from CBSE Class 9 Mathematics syllabus.

## RD Sharma Solutions Class 9 Maths Chapter 5 – Factorization Of Algebraic Expressions: PDF

RD Sharma Solutions Class 9 Maths Chapter 5

## Access answers of RD Sharma Solutions Class 9 Maths Chapter 5 : Factorization Of Algebraic Expressions

### RD Sharma Class 9 Chapter 5 Factorisation of Algebraic Expressions Ex 5.1

Factorize
Question 1.
x3 + x – 3x2 – 3
Solution:
x3 + x – 3x2 – 3
x3 – 3a2 + x – 3
⇒  x2(x – 3) + 1(x – 3)
= (x – 3) (x2 + 1)

Question 2.
a(a + b)3 – 3a2b(a + b)
Solution:
a(a + b)3 – 3a2b(a + b)
= a(a + b) {(a + b)2 – 3ab}
= a(a + b) {a2 + b2 + 2ab – 3ab}
= a{a + b) {a2 – ab + b2)

Question 3.
x(x3 – y3) + 3xy(x – y)
Solution:
x(x3 – y3) + 3xy(x – y)
= x(x – y) (x2 + xy + y2) + 3xy(x – y)
= x(x – y) (x2 + xy + y2 + 3y)
= x(x – y) (x2 + xy + y2 + 3y)

Question 4.
a2x2 + (ax2 +1)x + a
Solution:
a2x2 + (ax2 + 1)x + a
= a2x2 + a + (ax2 + 1)x
= a(ax2 + 1) + x(ax2 + 1)
= (ax2 + 1) (a + x)
= (x + a) (ax2 + 1)

Question 5.
x2 + y – xy – x
Solution:
x2 + y – xy – x
= x2-x-xy + y = x(x- l)-y(*- 1)
= (x – 1) (x – y)

Question 6.
X3 – 2x2y + 3xy2 – 6y3
Solution:
x3 – 2x2y + 3xy2 – 6y3
= x2(x – 2y)
+ 3y2(x – 2y)
= (x – 2y) (x2 + 3y2)

Question 7.
6ab – b2 + 12ac – 2bc
Solution:
6ab – b2 + 12ac – 2bc
= 6ab + 12ac – b2 – 2bc
= 6a(b + 2c) – b(b + 2c)
= (b + 2c) (6a – b)

Question 8.
x(x – 2) (x – 4) + 4x – 8
Solution:
x(x – 2) (x – 4) + 4x – 8
= x(x – 2) (x – 4) + 4(x – 2)
= (x – 2) [x(x – 4) + 4]
= (x – 2) (x2 – 4x + 4)
= (x – 2) [(x)2 – 2 x x x 2 + (2)2]
= (x – 2) (x – 2)2 = (x – 2)3

Question 9.
(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)      {∵ a2 + b2 + 2ab = (a + b)2}
= [a – b + c + b- c + a]2
= (2a)2 = 4a2

Question 10.
a2 + 2ab + b2 – c2
Solution:
a2 + 2ab + b2 – c2
= (a2 + 2ab + b2) – c2
= (a + b)2 – (c)2         {∵  a2 – b2 = (a + b) (a – b)}
= (a + b + c) (a + b – c)

Question 11.
a2 + 4b2 – 4ab – 4c2
Solution: Question 12.
x2 – y2 – 4xz + 4z2
Solution:
x2 – y2 – 4xz + 4z2
= x2 – 4xz + 4z2 – y2
= (x)2 – 2 x x x 2z + (2z)2 – (y)2
= (x – 2z)2 – (y)2
= (x – 2z + y) (x – 2z – y)
= (x +y – 2z) (x – y – 2z)

Question 13. Solution: Question 14. Solution: Question 15. Solution: Question 16.
Give possible expression for the length and breadth of the rectangle having 35y2 + 13y – 12 as its area.
Solution:
Area of a rectangle = 35y2 + 13y – 12
= 35y2 + 28y- 15y- 12 (i) If length = 5y + 4, then breadth = 7y – 3
(ii) and if length = 7y-3, then length = 5y+ 4

Question 17.
What are the possible expressions for the dimensions of the cuboid whose volume is 3x2 – 12x.
Solution:
Volume 3x2 – 12x
= 3x(x – 4)
∴ Factors are 3, x, and x – 4
Now, if length = 3, breadth = x and height = x – 4
if length =3, breadth = x – 4, height = x
if length = x, breadth = 3, height = x – 4
if length = x, breadth = x – 4, height = 3
if length = x – 4, breadth = 3, height = x
if length – x – 4, breadth = x, height = 3

Question 18. Solution: Question 19.
(x + 2) (x2 + 25) – 10x2 – 20x
Solution:
(x + 2) (x2 + 25) – 10x2 – 20x
= (x + 2) (x2 + 25) – 10x(x + 2)
= (x + 2) [x2 + 25 – 10x]
= (x + 2) [(x)2 – 2 x x x 5 + (5)2]
= (x + 2) (x – 5)2

Question 20.
2a2 + 26–√ ab +3b2
Solution:
2a2 + 26–√  ab +3 b2
= (2–√ a)22–√ a x 3–√ b+ (3–√ b)2
= (2–√a + 3–√ b)2

Question 21.
a2 + b2 + 2(ab + bc + ca)
Solution:
a2 + b2 + 2(ab + bc + ca)
= a2 + b2 + 2 ab + 2 bc + 2 ca
= (a + b)2 + 2c(b + a)
= (a + b)2 + 2c(a + b)
= (a + b) (a + b + 2c)

Question 22.
4(x – y)2 – 12(x -y) (x + y) + 9(x + y)2
Solution:
4(x – y)2 – 12(x – y) (x + y) + 9(x + y)2
= [2(x – y)2 + 2 x 2(x – y) x 3(x + y) + [3 (x+y]2        { a2 + b2 + 2 abc = (a + b)2}
= [2(x – y) + 3(x + y)]2
= (2x-2y + 3x + 3y)2
= (5x + y)2

Question 23.
a2 – b2 + 2bc – c2
Solution:
a2 – b2 + 2bc – c2
= a2 – (b2 – 2bc + c2)                                           { a2 + b2 – 2abc = (a – b)2}
= a2 – (b – c)2
= (a)2 – (b – c)2          { a2 – b2 = (a + b) (a – b)}
= (a + b – c) (a – b + c)

Question 24.
xy9 – yx9
Solution:
xy9 – yx9 = xy(y8 – x8)
= -xy(x8 – y8)
= -xy[(x4)2 – (y4)2]
= -xy (x4 + y4) (x4 – y4)                                         {∵ a2-b2 = (a + b) (a – b)}
= -xy (x4 + y4) {(x2)2 – (y2)2}
= -xy(x+ y4) (x2 + y2) (x2 – y2)
= -xy (x4 +y4) (x2 + y2) (x + y) (x -y)
= -xy(x – y) (x + y) (x2 + y2) (x4 + y4)

Question 25.
x4 + x2y2 + y4
Solution:
x4 + x2y2 + y4  = (x2)2 + 2x2y2 + y4 – x2y2           (Adding and subtracting x2y2)
= (x2 + y2)2 – (xy)2                                                        { a2 – b2 = (a + b) (a – b)}
= (x2 + y2 + xy) (x2 + y2 – xy)
= (x2 + xy + y2) (x2 – xy + y2)

Question 26.
x2 + 62–√x + 10
Solution: Question 27.
x2 + 22–√x- 30
Solution: Question 28.
x2 – 3–√x – 6
Solution: Question 29.
x2 + 5 5–√x + 30
Solution: Question 30.
x2 + 2 3–√x – 24
Solution: Question 31.
5–√x2 + 20x + 35–√
Solution:  Question 32.
2x2 + 35–√ x + 5
Solution: Question 33.
9(2a – b)2 – 4(2a – b) – 13
Solution: Question 34.
7(x-2y) – 25(x-2y) +12
Solution: Question 35.
2(x+y) – 9(x+y) -5
Solution:
2(x+y) – 9(x+y) -5 ### Factorisation of Algebraic Expressions RD Sharma Class 9 Solutions Chapter 5 Exercise-5.1

Factorisation of Algebraic Expressions RD Sharma Class 9 Solutions Chapter 5 Exercise-5.1 Q 1.                                   ## Important Topics Class 9 Maths Chapter 5

Well, nothing beats a quick revision session then getting through the important sections of the chapter. So, here we are with –

• Factorisation by taking out the common factors
• Factorisation by grouping the terms
• Factorisation by making a perfect square
• Factorisation by the difference of two squares
• The Factorisation of quadratic polynomials by splitting the middle term
• The Factorisation of algebraic expressions expressible as the sum or difference of two cubes

Well, this is everything that one needs to finish their CBSE Class 9 maths syllabus. The entire chapter simplifies the concept of factorisation and how to simplify an algebraic expression using the factorisation method. The process of factorization can be stated as the disintegration of a term into smaller factors. Whereas, the algebraic expressions are built up of variables, integer constants, and basic arithmetic operations of algebra.