Factorisation of Algebraic Expressions RD Sharma Class 9 Solutions

Factorisation of Algebraic Expressions RD Sharma Class 9 Solutions

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

Factorize
Question 1.
x ³ + x – 3x ² – 3
Solution:
x ³ + x – 3x ² – 3
x ³ – 3a ² + x – 3
⇒  x ² (x – 3) + 1(x – 3)
= (x – 3) (x ² + 1)

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

Question 3.
x(x ³ – y ³ ) + 3xy(x – y)
Solution:
x(x ³ – y ³ ) + 3xy(x – y)
= x(x – y) (x ² + xy + y ² ) + 3xy(x – y)
= x(x – y) (x ² + xy + y ² + 3y)
= x(x – y) (x ² + xy + y ² + 3y)

Question 4.
a ² x ² + (ax ² +1)x + a
Solution:
a ² x ² + (ax ² + 1)x + a
= a ² x ² + a + (ax ² + 1)x
= a(ax ² + 1) + x(ax ² + 1)
= (ax ² + 1) (a + x)
= (x + a) (ax ² + 1)

Question 5.
x ² + y – xy – x
Solution:
x ² + y – xy – x
= x ² -x-xy + y = x(x- l)-y(*- 1)
= (x – 1) (x – y)

Question 6.
X ³ – 2x ² y + 3xy ² – 6y ³
Solution:
x ³ – 2x ² y + 3xy ² – 6 y ³
= x ² (x – 2y) + 3y ² (x – 2y)
= (x – 2y) (x ² + 3y ² )

Question 7.
6 ab – b ² + 12ac – 2bc
Solution:
6ab – b ² + 12ac – 2bc
= 6ab + 12ac – b ² – 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) (x ² – 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) {∵ a ² + b ² + 2ab = (a + b) ² }
= [a – b + c + b- c + a] ²
= (2a) ² = 4a ²

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

Question 11.
a ² + 4b ² – 4ab – 4c ²
Solution:

Question 12.
x ² – y ² – 4xz + 4z ²
Solution:
x ² – y ² – 4xz + 4z ²
= x ² – 4xz + 4z ² – y ²
= (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 35y ² + 13y – 12 as its area.
Solution:
Area of a rectangle = 35y ² + 13y – 12
= 35y ² + 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 3x ² – 12x.
Solution:
Volume 3x ² – 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) (x ² + 25) – 10x ² – 20x
Solution:
(x + 2) (x ² + 25) – 10x ² – 20x
= (x + 2) (x ² + 25) – 10x(x + 2)
= (x + 2) [x ² + 25 – 10x]
= (x + 2) [(x) ² – 2 x x x 5 + (5) ² ]
= (x + 2) (x – 5) ²

Question 20.
2a ² + 2\(\sqrt { 6 } \) ab +3b ²
Solution:
2a ² + 2\(\sqrt { 6 } \)  ab +3 b ²
= (\(\sqrt { 2 } \) a) ² + \(\sqrt { 2 } \) a x \(\sqrt { 3 } \) b+ (\(\sqrt { 3 } \) b) ²
= (\(\sqrt { 2 } \)a + \(\sqrt { 3 } \) b) 2

Question 21.
a ² + b ² + 2(ab + bc + ca)
Solution:
a ² + b ² + 2(ab + bc + ca)
= a ² + b ² + 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 { ∵ a ² + b ² + 2 abc = (a + b) ² }
= [2(x – y) + 3(x + y)] ²
= (2x-2y + 3x + 3y) ²
= (5x + y) ²

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

Question 24.
xy ⁹ – yx ⁹
Solution:
xy ⁹ – yx ⁹ = xy(y ⁸ – x ⁸ )
= -xy(x ⁸ – y ⁸ )
= -xy[(x ⁴ ) ² – (y ⁴ ) ² ]
= -xy (x ⁴ + y ⁴ ) (x ⁴ – y ⁴ )                                         { ∵ a ² -b ² = (a + b) (a – b)}
= -xy (x ⁴ + y ⁴ ) {(x ² ) ² – (y ² ) ² }
= -xy(x ⁴ + y ⁴ ) (x ² + y ² ) (x ² – y ² )
= -xy (x ⁴ +y ⁴ ) (x ² + y ² ) (x + y) (x -y)
= -xy(x – y) (x + y) (x ² + y ² ) (x ⁴ + y ⁴ )

Question 25.
x ⁴ + x ² y ² + y ⁴
Solution:
x ⁴ + x ² y ² + y ⁴ = (x ² ) ² + 2x ² y ² + y ⁴ – x ² y ² (Adding and subtracting x ² y ² )
= (x ² + y ² ) ² – (xy) ² { ∵ a ² – b ² = (a + b) (a – b)}
= (x ² + y ² + xy) (x ² + y ² – xy)
= (x ² + xy + y ² ) (x ² – xy + y ² )

Question 26.
x ² + 6\(\sqrt { 2 } \)x + 10
Solution:

Question 27.
x ² + 2\(\sqrt { 2 } \)x- 30
Solution:

Question 28.
x ² – \(\sqrt { 3 } \)x – 6
Solution:

Question 29.
x ² + 5 \(\sqrt { 5 } \)x + 30
Solution:

Question 30.
x ² + 2 \(\sqrt { 3 } \)x – 24
Solution:

Question 31.
5 \(\sqrt { 5 } \)x ² + 20x + 3\(\sqrt { 5 } \)
Solution:

Question 32.
2x ² + 3\(\sqrt { 5 } \) 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.

• Factorisation of Algebraic Expressions Exercise 5.1
• Factorisation of Algebraic Expressions Exercise 5.2
• Factorisation of Algebraic Expressions Exercise 5.3
• Factorisation of Algebraic Expressions Exercise 5.4

RD Sharma Class 9 Solutions