RD Sharma Solutions Class 9 Maths Chapter 5 Factorisation of Algebraic Expressions
Here we have provided RD Sharma Class 9 Solutions Maths Chapter 5 solutions for the students to help them ace their examinations. Students can refer to these solutions and practice these questions to score better in the exams.
RD Sharma Solutions Class 9 Maths Chapter 5 PDF
RD Sharma Solutions Class 9 Maths Chapter 5 Factorisation of Algebraic Expressions
RD Sharma Solutions Class 9 Maths Chapter 5 Factorisation of Algebraic Expressions Exercise 5.1 Page No: 5.9
Question 1: Factorize x
3
+ x – 3x
2
– 3
Solution:
x
3
+ x – 3x
2
– 3
Here x is common factor in x
3
+ x and – 3 is common factor in – 3x
2
– 3
x
3
– 3x
2
+ x – 3
x
2
(x – 3) + 1(x – 3)
Taking ( x – 3) common
(x – 3) (x
2
+ 1)
Therefore x
3
+ x – 3x
2
– 3 = (x – 3) (x
2
+ 1)
Question 2: Factorize a(a + b)
3
– 3a
2
b(a + b)
Solution
:
a(a + b)
3
– 3a
2
b(a + b)
Taking a (a + b) as common factor
= a(a + b) {(a + b)
2
– 3ab}
= a(a + b) {a
2
+ b
2
+ 2ab – 3ab}
= a(a + b) (a
2
+ b
2
– ab)
Question 3: Factorize 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)
Taking x(x – y) as a common factor
= x(x – y) (x
2
+ xy + y
2
+ 3y)
= x(x – y) (x
2
+ xy + y
2
+ 3y)
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Question 4: Factorize 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)
Question 5: Factorize x
2
+ y – xy – x
Solution:
x
2
+ y – xy – x
= x
2
– x – xy + y
= x(x- 1) – y(x – 1)
= (x – 1) (x – y)
Question 6: Factorize x
3
– 2x
2
y + 3xy
2
– 6y
3
Solution:
x
3
– 2x
2
y + 3xy
2
– 6y
3
= x
2
(x – 2y) + 3y
2
(x – 2y)
= (x – 2y) (x
2
+ 3y
2
)
Question 7: Factorize 6ab – b
2
+ 12ac – 2bc
Solution:
6ab – b
2
+ 12ac – 2bc
= 6ab + 12ac – b
2
– 2bc
Taking 6a common from first two terms and –b from last two terms
= 6a(b + 2c) – b(b + 2c)
Taking (b + 2c) common factor
= (b + 2c) (6a – b)
Question 8: Factorize (x
2
+ 1/x
2
) – 4(x + 1/x) + 6
Solution:
(x
2
+ 1/x
2
) – 4(x + 1/x) + 6
= x
2
+ 1/x
2
– 4x – 4/x + 4 + 2
= x
2
+ 1/x
2
+ 4 + 2 – 4/x – 4x
= (x
2
) + (1/x)
2
+ ( -2 )
2
+ 2x(1/x) + 2(1/x)(-2) + 2(-2)x
As we know, x
2
+ y
2
+ z
2
+ 2xy + 2yz + 2zx = (x+y+z)
2
So, we can write;
= (x + 1/x + (-2 ))
2
or (x + 1/x – 2)
2
Therefore, x
2
+ 1/x
2
) – 4(x + 1/x) + 6 = (x + 1/x – 2)
2
Question 9: Factorize 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)(2) + (2)
2
]
= (x – 2) (x – 2)
2
= (x – 2)
3
Question 10: Factorize ( x + 2 ) ( x
2
+ 25 ) – 10x
2
– 20x
Solution :
( x + 2) ( x
2
+ 25) – 10x ( x + 2 )
Take ( x + 2 ) as common factor;
= ( x + 2 )( x
2
+ 25 – 10x)
=( x + 2 ) ( x
2
– 10x + 25)
Expanding the middle term of ( x
2
– 10x + 25 )
=( x + 2 ) ( x
2
– 5x – 5x + 25 )
=( x + 2 ){ x (x – 5 ) – 5 ( x – 5 )}
=( x + 2 )( x – 5 )( x – 5 )
=( x + 2 )( x – 5 )
2
Therefore, ( x + 2) ( x
2
+ 25) – 10x ( x + 2 ) = ( x + 2 )( x – 5 )
2
Question 11: Factorize 2a
2
+ 2√6 ab + 3b
2
Solution:
2a
2
+ 2√6 ab + 3b
2
Above expression can be written as ( √2a )
2
+ 2 × √2a × √3b + ( √3b)
2
As we know, ( p + q )
2
= p
2
+ q
2
+ 2pq
Here p = √2a and q = √3b
= (√2a + √3b )
2
Therefore, 2a
2
+ 2√6 ab + 3b
2
= (√2a + √3b )
2
Question 12: Factorize (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 p
2
+ q
2
+ 2pq = (p + q)
2
}
Here p = a – b + c and q = b – c + a
= [a – b + c + b- c + a]
2
= (2a)
2
= 4a
2
Question 13: Factorize a
2
+ b
2
+ 2( ab+bc+ca )
Solution
:
a
2
+ b
2
+ 2ab + 2bc + 2ca
As we know, p
2
+ q
2
+ 2pq = (p + q)
2
We get,
= ( a+b)
2
+ 2bc + 2ca
= ( a+b)
2
+ 2c( b + a )
Or ( a+b)
2
+ 2c( a + b )
Take ( a + b ) as common factor;
= ( a + b )( a + b + 2c )
Therefore, a
2
+ b
2
+ 2ab + 2bc + 2ca = ( a + b )( a + b + 2c )
Question 14: Factorize 4(x-y)
2
– 12(x – y)(x + y) + 9(x + y)
2
Solution :
Consider ( x – y ) = p, ( x + y ) = q
= 4p
2
– 12pq + 9q
2
Expanding the middle term, -12 = -6 -6 also 4× 9=-6 × -6
= 4p
2
– 6pq – 6pq + 9q
2
=2p( 2p – 3q ) -3q( 2p – 3q )
= ( 2p – 3q ) ( 2p – 3q )
= ( 2p – 3q )
2
Substituting back p = x – y and q = x + y;
= [2( x-y ) – 3( x+y)]
2
= [ 2x – 2y – 3x – 3y ]
2
= (2x-3x-2y-3y )
2
=[ -x – 5y]
2
=[( -1 )( x+5y )]
2
=( x+5y )
2
Therefore, 4(x-y)
2
– 12(x – y)(x + y) + 9(x + y)
2
= ( x+5y )
2
Question 15: Factorize a
2
– b
2
+ 2bc – c
2
Solution :
a
2
– b
2
+ 2bc – c
2
As we know, ( a-b)
2
= a
2
+ b
2
– 2ab
= a
2
– ( b – c)
2
Also we know, a
2
– b
2
= ( a+b)( a-b)
= ( a + b – c )( a – ( b – c ))
= ( a + b – c )( a – b + c )
Therefore, a
2
– b
2
+ 2bc – c
2
=( a + b – c )( a – b + c )
Question 16: Factorize 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
We know, a
2
– b
2
= (a + b) (a – b)
= (a + b + c) (a + b – c)
Therefore a
2
+ 2ab + b
2
– c
2
= (a + b + c) (a + b – c)
RD Sharma Solutions Class 9 Maths Chapter 5 Factorisation of Algebraic Expressions Exercise 5.2 Page No: 5.13
Factorize each of the following expressions:
Question 1: p
3
+ 27
Solution
:
p
3
+ 27
= p
3
+ 3
3
[using a
3
+ b
3
= (a + b)(a
2
–ab + b
2
)]
= (p + 3)(p² – 3p – 9)
Therefore, p
3
+ 27 = (p + 3)(p² – 3p – 9)
Question 2: y
3
+ 125
Solution
:
y
3
+ 125
= y
3
+ 5
3
[using a
3
+ b
3
= (a + b)(a
2
–ab + b
2
)]
= (y+5)(y
2
− 5y + 5
2
)
= (y + 5)(y
2
− 5y + 25)
Therefore, y
3
+ 125 = (y + 5)(y
2
− 5y + 25)
Question 3: 1 – 27a
3
Solution
:
= (1)
3
−(3a)
3
[using a
3
– b
3
= (a – b)(a
2
+ ab + b
2
)]
= (1− 3a)(1
2
+ 1×3a + (3a)
2
)
= (1−3a)(1 + 3a + 9a
2
)
Therefore, 1−27a
3
= (1−3a)(1 + 3a+ 9a
2
)
Question 4: 8x
3
y
3
+ 27a
3
Solution:
8x
3
y
3
+ 27a
3
= (2xy)
3
+ (3a)
3
[using a
3
+ b
3
= (a + b)(a
2
–ab + b
2
)]
= (2xy +3a)((2xy)
2
−2xy×3a+(3a)
2
)
= (2xy+3a)(4x
2
y
2
−6xya + 9a
2
)
Question 5: 64a
3
− b
3
Solution:
64a
3
− b
3
= (4a)
3
−b
3
[using a
3
– b
3
= (a – b)(a
2
+ ab + b
2
)]
= (4a−b)((4a)
2
+ 4a×b + b
2
)
=(4a−b)(16a
2
+4ab+b
2
)
Question 6: x
3
/ 216 – 8y
3
Solution:
x
3
/ 216 – 8y
3
Question 7: 10x
4
y – 10xy
4
Solution:
10x
4
y – 10xy
4
= 10xy(x
3
− y
3
)
[using a
3
– b
3
= (a – b)(a
2
+ ab + b
2
)]
= 10xy (x−y)(x
2
+ xy + y
2
)
Therefore, 10x
4
y – 10xy
4
= 10xy (x−y)(x
2
+ xy + y
2
)
Question 8: 54x
6
y + 2x
3
y
4
Solution:
54x
6
y + 2x
3
y
4
= 2x
3
y(27x
3
+y
3
)
= 2x
3
y((3x)
3
+ y
3
)
[using a
3
+ b
3
= (a + b)(a
2
– ab + b
2
)]
= 2x
3
y {(3x+y) ((3x)
2
−3xy+y
2
)}
=2x
3
y(3x+y)(9x
2
− 3xy + y
2
)
Question 9: 32a
3
+ 108b
3
Solution:
32a
3
+ 108b
3
= 4(8a
3
+ 27b
3
)
= 4((2a)
3
+(3b)
3
)
[using a
3
+ b
3
= (a + b)(a
2
– ab + b
2
)]
= 4[(2a+3b)((2a)
2
−2a×3b+(3b)
2
)]
= 4(2a+3b)(4a
2
− 6ab + 9b
2
)
Question 10: (a−2b)
3
− 512b
3
Solution:
(a−2b)
3
− 512b
3
= (a−2b)
3
−(8b)
3
[using a
3
– b
3
= (a – b)(a
2
+ ab + b
2
)]
= (a −2b−8b) {(a−2b)
2
+ (a−2b)8b + (8b)
2
}
=(a −10b)(a
2
+ 4b
2
− 4ab + 8ab − 16b
2
+ 64b
2
)
=(a−10b)(a
2
+ 52b
2
+ 4ab)
Question 11: (a+b)
3
− 8(a−b)
3
Solution:
(a+b)
3
− 8(a−b)
3
= (a+b)
3
− [2(a−b)]
3
= (a+b)
3
− [2a−2b]
3
[using p
3
– q
3
= (p – q)(p
2
+ pq + q
2
)]
Here p = a+b and q = 2a−2b
= (a+b−(2a−2b))((a+b)
2
+(a+b)(2a−2b)+(2a−2b)
2
)
=(a+b−2a+2b)(a
2
+b
2
+2ab+(a+b)(2a−2b)+(2a−2b)
2
)
=(a+b−2a+2b)(a
2
+b
2
+2ab+2a
2
−2ab+2ab−2b
2
+(2a−2b)
2
)
=(3b−a)(3a
2
+2ab−b
2
+(2a−2b)
2
)
=(3b−a)(3a
2
+2ab−b
2
+4a
2
+4b
2
−8ab)
=(3b−a)(3a
2
+4a
2
−b
2
+4b
2
−8ab+2ab)
=(3b−a)(7a
2
+3b
2
−6ab)
Question 12: (x+2)
3
+ (x−2)
3
Solution:
(x+2)
3
+ (x−2)
3
[using p
3
+ q
3
= (p + q)(p
2
– pq + q
2
)]
Here p = x + 2 and q = x – 2
= (x+2+x−2)((x+2)
2
−(x+2)(x−2)+(x−2)
2
)
=2x(x
2
+4x+4−(x+2)(x−2)+x
2
−4x+4)
[ Using : (a+b)(a−b) = a
2
−b
2
]
= 2x(2x
2
+ 8 − (x
2
− 2
2
))
= 2x(2x
2
+8 − x
2
+ 4)
= 2x(x
2
+ 12)
RD Sharma Solutions Class 9 Maths Chapter 5 Factorisation of Algebraic Expressions Exercise 5.3 Page No: 5.17
Question 1: Factorize 64a
3
+ 125b
3
+ 240a
2
b + 300ab
2
Solution:
64a
3
+ 125b
3
+ 240a
2
b + 300ab
2
= (4a)
3
+ (5b)
3
+ 3(4a)
2
(5b) + 3(4a)(5b)
2
, which is similar to a
3
+ b
3
+ 3a
2
b + 3ab
2
We know that, a
3
+ b
3
+ 3a
2
b + 3ab
2
= (a+b)
3
]
= (4a+5b)
3
Question 2: Factorize 125x
3
– 27y
3
– 225x
2
y + 135xy
2
Solution
:
125x
3
– 27y
3
– 225x
2
y + 135xy
2
Above expression can be written as (5x)
3
−(3y)
3
−3(5x)
2
(3y) + 3(5x)(3y)
2
Using: a
3
− b
3
− 3a
2
b + 3ab
2
= (a−b)
3
= (5x − 3y)
3
Question 3: Factorize 8/27 x
3
+ 1 + 4/3 x
2
+ 2x
Solution
:
8/27 x
3
+ 1 + 4/3 x
2
+ 2x
Question 4: Factorize 8x
3
+ 27y
3
+ 36x
2
y + 54xy
2
Solution:
8x
3
+ 27y
3
+ 36x
2
y + 54xy
2
Above expression can be written as (2x)
3
+ (3y)
3
+ 3×(2x)
2
×3y + 3×(2x)(3y)
2
Which is similar to a³ + b³ + 3a²b + 3ab² = (a + b) ³]
Here a = 2x and b = 3y
= (2x+3y)
3
Therefore, 8x
3
+ 27y
3
+ 36x
2
y + 54xy
2
= (2x+3y)
3
Question 5: Factorize a
3
− 3a
2
b + 3ab
2
− b
3
+ 8
Solution
:
a
3
− 3a
2
b + 3ab
2
− b
3
+ 8
Using: a
3
− b
3
− 3a
2
b + 3ab
2
= (a−b)
3
= (a−b)
3
+ 2
3
Again ,
Using: a
3
+ b
3
=(a + b)(a
2
– ab + b
2
)]
=(a−b+2)((a−b)
2
−(a−b) × 2 + 2
2
)
=(a−b+2)(a
2
+b
2
−2ab−2(a−b)+4)
=(a−b+2)(a
2
+b
2
−2ab−2a+2b+4)
a
3
− 3a
2
b + 3ab
2
− b
3
+ 8 =(a−b+2)(a
2
+b
2
−2ab−2a+2b+4)
RD Sharma Solutions Class 9 Maths Chapter 5 Factorisation of Algebraic Expressions Exercise 5.4 Page No: 5.22
Factorize each of the following expressions:
Question 1: a
3
+ 8b
3
+ 64c
3
− 24abc
Solution
:
a
3
+ 8b
3
+ 64c
3
− 24abc
= (a)
3
+ (2b)
3
+ (4c)
3
− 3×a×2b×4c
[Using a
3
+b
3
+c
3
−3abc = (a+b+c)(a
2
+b
2
+c
2
−ab−bc−ca)]
= (a+2b+4c)(a
2
+(2b)
2
+ (4c)
2
−a×2b−2b×4c−4c×a)
= (a+2b+4c)(a
2
+4b
2
+16c
2
−2ab−8bc−4ac)
Therefore, a
3
+ 8b
3
+ 64c
3
− 24abc = (a+2b+4c)(a
2
+4b
2
+16c
2
−2ab−8bc−4ac)
Question 2: x
3
− 8y
3
+ 27z
3
+ 18xyz
Solution:
= x
3
− (2y)
3
+ (3z)
3
− 3×x×(−2y)(3z)
= (x + (−2y) + 3z) (x
2
+ (−2y)
2
+ (3z)
2
−x(−2y)−(−2y)(3z)−3z(x))
[using a
3
+b
3
+c
3
−3abc = (a+b+c)(a
2
+b
2
+c
2
−ab−bc−ca)]
=(x −2y + 3z)(x
2
+ 4y
2
+ 9 z
2
+ 2xy + 6yz − 3zx)
Question 3: 27x
3
− y
3
– z
3
– 9xyz
Solution
:
27x
3
− y
3
– z
3
– 9xyz
= (3x)
3
− y
3
– z
3
– 3(3xyz)
[Using a
3
+ b
3
+ c
3
−3abc = (a + b + c)(a
2
+b
2
+c
2
−ab−bc−ca)]
Here a = 3x, b = -y and c = -z
= (3x – y – z){ (3x)
2
+ (- y)
2
+ (– z)
2
+ 3xy – yz + 3xz)}
= (3x – y – z){ 9x
2
+ y
2
+ z
2
+ 3xy – yz + 3xz)}
Question 4: 1/27 x
3
− y
3
+ 125z
3
+ 5xyz
Solution:
1/27 x
3
− y
3
+ 125z
3
+ 5xyz
= (x/3)
3
+(−y)
3
+(5z)
3
– 3 x/3 (−y)(5z)
[Using a
3
+ b
3
+ c
3
−3abc = (a + b + c)(a
2
+b
2
+c
2
−ab−bc−ca)]
= (x/3 + (−y) + 5z)((x/3)
2
+ (−y)
2
+ (5z)
2
–x/3(−y) − (−y)5z−5z(x/3))
= (x/3 −y + 5z) (x^2/9 + y
2
+ 25z
2
+ xy/3 + 5yz – 5zx/3)
Question 5: 8x
3
+ 27y
3
− 216z
3
+ 108xyz
Solution
:
8x
3
+ 27y
3
− 216z
3
+ 108xyz
= (2x)
3
+ (3y)
3
+(−6y)
3
−3(2x)(3y)(−6z)
= (2x+3y+(−6z)){ (2x)
2
+(3y)
2
+(−6z)
2
−2x×3y−3y(−6z)−(−6z)2x}
= (2x+3y−6z) {4x
2
+9y
2
+36z
2
−6xy + 18yz + 12zx}
Question 6: 125 + 8x
3
− 27y
3
+ 90xy
Solution:
125 + 8x
3
− 27y
3
+ 90xy
= (5)
3
+ (2x)
3
+(−3y)
3
−3×5×2x×(−3y)
= (5+2x+(−3y)) (5
2
+(2x)
2
+(−3y)
2
−5(2x)−2x(−3y)−(−3y)5)
= (5+2x−3y)(25+4x
2
+9y
2
−10x+6xy+15y)
Question 7: (3x−2y)
3
+ (2y−4z)
3
+ (4z−3x)
3
Solution:
(3x−2y)
3
+ (2y−4z)
3
+ (4z−3x)
3
Let (3x−2y) = a, (2y−4z) = b , (4z−3x) = c
a + b + c= 3x−2y+2y−4z+4z−3x = 0
We know, a
3
+ b
3
+ c
3
−3abc = (a + b + c)(a
2
+b
2
+c
2
−ab−bc−ca)
⇒ a
3
+ b
3
+ c
3
−3abc = 0
or a
3
+ b
3
+ c
3
=3abc
⇒ (3x−2y)
3
+ (2y−4z)
3
+ (4z−3x)
3
= 3(3x−2y)(2y−4z)(4z−3x)
Question 8: (2x−3y)
3
+ (4z−2x)
3
+ (3y−4z)
3
Solution:
(2x−3y)
3
+ (4z−2x)
3
+ (3y−4z)
3
Let 2x – 3y = a , 4z – 2x = b , 3y – 4z = c
a + b + c= 2x – 3y + 4z – 2x + 3y – 4z = 0
We know, a
3
+ b
3
+ c
3
−3abc = (a + b + c)(a
2
+b
2
+c
2
−ab−bc−ca)
⇒ a
3
+ b
3
+ c
3
−3abc = 0
(2x−3y)
3
+ (4z−2x)
3
+ (3y−4z)
3
= 3(2x−3y)(4z−2x)(3y−4z)
RD Sharma Solutions Class 9 Maths Chapter 5 Factorisation of Algebraic Expressions Exercise VSAQs Page No: 5.24
Question 1: Factorize x
4
+ x
2
+ 25
Solution:
x
4
+ x
2
+ 25
= (x
2
)
2
+ 5
2
+ x
2
[using a
2
+ b
2
= (a + b)
2
– 2ab ]
= (x
2
+5)
2
−2(x
2
) (5) + x
2
=(x
2
+5)
2
−10x
2
+ x
2
=(x
2
+ 5)
2
− 9x
2
=(x
2
+ 5)
2
− (3x)
2
[using a
2
– b
2
= (a + b)(a – b ]
= (x
2
+ 3x + 5)(x
2
− 3x + 5)
Question 2: Factorize x
2
– 1 – 2a – a
2
Solution:
x
2
– 1 – 2a – a
2
x
2
– (1 + 2a + a
2
)
x
2
– (a + 1)
2
(x – (a + 1)(x + (a + 1)
(x – a – 1)(x + a + 1)
[using a
2
– b
2
= (a + b)(a – b) and (a + b)^2 = a^2 + b^2 + 2ab ]
Question 3: If a + b + c =0, then write the value of a
3
+ b
3
+ c
3
.
Solution:
We know, a
3
+ b
3
+ c
3
– 3abc = (a + b +c ) (a
2
+ b
2
+ c
2
– ab – bc − ca)
Put a + b + c =0
This implies
a
3
+ b
3
+ c
3
= 3abc
Question 4: If a
2
+ b
2
+ c
2
= 20 and a + b + c =0, find ab + bc + ca.
Solution:
We know, (a+b+c)² = a² + b² + c² + 2(ab + bc + ca)
0 = 20 + 2(ab + bc + ca)
-10 = ab + bc + ca
Or
ab + bc + ca = -10
Question 5: If a + b + c = 9 and ab + bc + ca = 40, find a
2
+ b
2
+ c
2
.
Solution:
We know, (a+b+c)² = a² + b² + c² + 2(ab + bc + ca)
9
2
= a² + b² + c² + 2(40)
81 = a² + b² + c² + 80
⇒ a² + b² + c² = 1
