Factorisation ICSE Class-9th Concise Mathematics Selina Solutions Chapter-5. We provide step by step Solutions of Exercise / lesson-5 Factorisation for ICSE Class-9 Concise Selina Mathematics by RK Bansal.
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Factorisation ICSE Class-9th Concise Mathematics Selina Solutions Chapter-5
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Selina Solutions Exercise – 5(A), Factorisation for ICSE Class-9th Concise Mathematics
Question 1
Factorise by taking out the common factors :
2 (2x – 5y) (3x + 4y) – 6 (2x – 5y) (x – y)
Answer
2 (2x – 5y) (3x + 4y) – 6 (2x – 5y) (x – y)
Taking (2x – 5y) common from both terms
= (2x – 5y)[2(3x + 4y) – 6(x – y)]
= (2x – 5y)(6x + 8y – 6x + 6y)
= (2x – 5y)(8y + 6y)
= (2x – 5y)(14y)
= (2x – 5y)14y
Question 2
Factories by taking out common factors :
xy(3x2 – 2y2) – yz(2y2 – 3x2) + zx(15x2 – 10y2)
Answer
xy(3x2 – 2y2) – yz(2y2 – 3x2) + zx(15x2 – 10y2)
= xy(3x2 – 2y2) + yz(3x2 – 2y2) + zx(15x2 – 10y2)
= xy(3x2 – 2y2) + yz(3x2 – 2y2) + 5zx(3x2 – 2y2)
= (3x2 – 2y2)[xy + yz + 5zx]
Question 3
Factories by taking out common factors :
ab(a2 + b2 – c2) – bc(c2 – a2 – b2) + ca(a2 + b2 – c2)
Answer
ab(a2 + b2 – c2) – bc(c2 – a2 – b2) + ca(a2 + b2 – c2)
= ab(a2 + b2 – c2) + bc(a2 + b2 – c2) + ca(a2 + b2 – c2)
= (a2 + b2 – c2)[ab + bc + ca]
Question 4
Factories by taking out common factors :
2x(a – b) + 3y(5a – 5b) + 4z(2b – 2a)
Answer
2x(a – b) + 3y(5a – 5b) + 4z(2b – 2a)
= 2x(a – b) + 15y(a – b) – 8z(a – b)
= (a – b)[2x + 15y – 8z]
Question 5
Factorise by the grouping method : a3 + a – 3a2 – 3
Answer
a3 + a – 3a2 – 3
= a (a2 + 1) – 3(a2 + 1)
= (a2 + 1) (a -3).
Question 6
Factorise by the grouping method: 16 (a + b)2 – 4a – 4b
Answer
16 (a + b)2 – 4a – 4b =16 (a + b)2 – 4 (a + b)
= 4 (a + b) [4 (a + b) – 1]
= 4 (a + b) (4a + 4b – 1)
Question 7
Factorise by the grouping method : a4 – 2a3 – 4a + 8
Answer
a4 – 2a3 – 4a + 8 = a3( a – 2 ) – 4( a – 2 )
= ( a3 – 4 )( a – 2 )
Question 8
Factorise by the grouping method : ab – 2b + a2 – 2a
Answer
ab – 2b + a2 – 2a = b( a – 2 ) + a( a – 2 )
= ( a + b )( a – 2 )
Question 9
Factorise by the grouping method : ab (x2 + 1) + x (a2 + b2)
Answer
ab (x2 + 1) + x (a2 + b2) = abx2 + ab + a2x + b2x
= ax( bx + a ) + b( bx + a )
= ( ax + b )( bx + a )
Question 10
Factorise by the grouping method : a2 + b – ab – a
Answer
a2 + b – ab – a = a2 – a + b – ab
= a( a – 1) + b( 1 – a )
= a(a – 1) – b(a – 1)
= (a -1)(a – b)
Question 11
Factorise by the grouping method : (ax + by)2 + (bx – ay)2
Answer
(ax + by)2 + (bx – ay)2
= a2x2 + b2y2 + 2axby + b2x2 + a2y2 – 2bxay
= a2x2 + b2y2 + b2x2 + a2y2
= x2( a2 + b2 ) + y2( a2 + b2 )
= ( x2 + y2 )( a2 + b2 )
Question 12
Factorise by the grouping method : a2x2 + (ax2 + 1) x + a
Answer
a2x2 + (ax2 + 1) x + a
= a2x2 + a + (ax2 + 1) x
= a( ax2 + 1) + x( ax2 + 1)
= ( a + x )( ax2 + 1 )
Question 13
Factorise by the grouping method : (2a-b)2 -10a + 5b
Answer
( 2a – b)2 – 10a + 5b
= ( 2a – b )2 – 5( 2a – b )
= ( 2a – b )( 2a – b – 5 )
Question 14
Factorise by the grouping method : a (a -4) – a + 4
Answer
a (a -4) – a + 4
= a( a – 4 ) -1( a – 4 )
= ( a – 4 )( a – 1 )
Question 15
Factorise by the grouping method : y2 – (a + b) y + ab
Answer
y2 – (a + b) y + ab
= y2 – ay – by + ab
= y( y – a ) – b( y – a )
= ( y – a )( y – b )
Question 16
Factorise by the grouping method :
Answer
Question 17
Factorise using the grouping method:
x2 + y2 + x + y + 2xy
Answer
x2 + y2 + x + y + 2xy
= ( x2 + y2 + 2xy ) + ( x + y ) [As (x + y)2 = x2 + 2xy + y2]
and = ( x + y )2 + ( x + y )
= ( x + y )( x + y + 1 )
Question 18
Factorise using the grouping method :
a2 + 4b2 – 3a + 6b – 4ab
Answer
a2 + 4b2 – 3a + 6b – 4ab
= a2 + 4b2 – 4ab – 3a + 6b
= a2 + (2b)2 – 2 × a × (2b) – 3(a – 2b) [As (a – b)2 = a2 – 2ab + b2 ]
= (a – 2b)2 – 3(a – 2b)
= (a – 2b)[(a – 2b)- 3]
= (a – 2b)(a – 2b – 3)
Question 19
Factorise using the grouping method :
m (x – 3y)2 + n (3y – x) + 5x – 15y
Answer
m (x – 3y)2 + n (3y – x) + 5x – 15y
= m (x – 3y)2 – n (x – 3y) + 5(x – 3y)
[Taking (x – 3y) common from all the three terms]
= (x – 3y) [m(x – 3y) – n + 5]
= (x – 3y)(mx – 3my – n + 5)
Question 20
Factorise using the grouping method :
x (6x – 5y) – 4 (6x – 5y)2
Answer
x (6x – 5y) – 4 (6x – 5y)2
= (6x – 5y)[x – 4(6x – 5y)]
[Taking (6x – 5y) common from the three terms]
= (6x – 5y)(x – 24x + 20y)
= (6x – 5y)(-23x + 20y)
= (6x – 5y)(20y – 23x)
Exe-5 B , Factorisation ICSE Class-9th Concise Mathematics Selina Solutions
Question 1
Factorise : a2 + 10a + 24
Answer
a2 + 10a + 24
= a2 + 6a + 4a + 24
= a( a + 6 ) + 4( a + 6 )
= ( a + 6 )( a + 4 )
Question 2
Factorise : a2 – 3a – 40
Answer
a2 – 3a – 40
= a2 – 8a + 5a – 40
= a( a – 8 ) + 5( a – 8 )
= ( a – 8 )( a + 5 )
Question 3
Factorise : 1 – 2a – 3a2
Answer
1 – 2a – 3a2
= 1 – 3a + a – 3a2
=( 1 + a )( 1 – 3a )
Question 4
Factorise : x2 – 3ax – 88a2
Answer
x2 – 3ax – 88a2
= x2 – 11ax + 8ax – 88a2
= x( x – 11a ) + 8a( x – 11a )
= ( x + 8a )( x – 11a )
Question 5
Factorise : 6a2 – a – 15
Answer
6a2 – a – 15
= 6a2 – 10a + 9a – 15
= 2a( 3a – 5 ) + 3( 3a – 5 )
= ( 2a + 3 )( 3a – 5 )
Question 6
Factorise : 24a3 + 37a2 – 5a
Answer
24a3 + 37a2 – 5a
= a( 24a2 + 37a – 5 )
= a( 24a2 + 40a -3a – 5 )
= a x [ 8a( 3a + 5 ) – 1( 3a + 5 )]
= a[( 8a – 1 )( 3a + 5 )]
= a( 8a – 1 )( 3a + 5 )
Question 7
Factorise : a(3a – 2) – 1
Answer
a(3a – 2) – 1
= 3a2 – 2a – 1
= 3a2 – 3a + a – 1
= 3a( a – 1 ) + 1( a – 1 )
= ( 3a + 1 )( a – 1 )
Question 8
Factorise : a2b2 + 8ab – 9
Answer
a2b2 + 8ab – 9
= a2b2 + 9ab – ab – 9
= ab( ab + 9 ) -1( ab + 9 )
= ( ab + 9 )( ab – 1 )
Question 9
Factorise : 3 – a (4 + 7a)
Answer
3 – a (4 + 7a)
= 3 – 4a – 7a2
= 3 – 7a + 3a – 7a2
= 1( 3 – 7a ) + a( 3 – 7a )
= ( 3 – 7a )( a + 1 )
Question 10
Factorise : (2a + b)2 – 6a – 3b – 4
Answer
(2a + b)2 – 6a – 3b – 4
= ( 2a + b )2 – 3( 2a + b ) – 4
Assume that, 2a + b = x
Therefore,
(2a + b)2 – 6a – 3b – 4
= x2 – 3x – 4
= x2 – 4x + x – 4
= 1( x – 4 ) + x( x – 4 )
= ( x + 1 )( x – 4 )
= ( 2a + b + 1 )( 2a + b – 4 )
[ resubstitute the value of x ]
Question 11
Factorise : 1 – 2 (a+ b) – 3 (a + b)2
Answer
1 – 2 (a+ b) – 3 (a + b)2
Assume that a + b = x ;
1 – 2( a + b ) – 3( a + b )2
= 1 – 2x – 3×2
= 1 – 3x + x – 3×2
= 1( 1 – 3x ) + x( 1 – 3x )
= ( 1 – 3x )( 1 + x )
= [ 1 – 3( a + b )][ 1 + ( a + b )]
= ( 1 – 3a – 3b )( 1 + a + b )
Question 12
Factorise : 3a2 – 1 – 2a
Answer
3a2 – 1 – 2a
= 3a2 – 2a – 1
= 3a2 – 3a + a – 1
= 3a( a – 1 ) + 1( a – 1 )
= ( 3a + 1 )( a – 1 )
Question 13
Factorise : x2 + 3x + 2 + ax + 2a
Answer
x2 + 3x + 2 + ax + 2a
= x2 + 2x + x + 2 + ax + 2a
= x( x + 2 )+1( x + 2 ) +a( x + 2 )
= ( x + 2 )( x + a + 1 )
Question 14
Factorise : (3x – 2y)2 + 3 (3x – 2y) – 10
Answer
(3x – 2y)2 + 3 (3x – 2y) – 10
Assume that 3x – 2y = a
Therefore,
(3x – 2y)2 + 3 (3x – 2y) – 10
= a2 + 3a – 10
= a2 + 5a – 2a -10
= a( a + 5 ) -2 ( a + 5 )
= ( a + 5 )( a – 2 )
= ( 3x – 2y + 5 )( 3x – 2y – 2)
Question 15
Factorise : 5 – (3a2 – 2a) (6 – 3a2 + 2a)
Answer
5 – (3a2 – 2a) (6 – 3a2 + 2a)
= 5 – ( 3a2 – 2a )[ 6 – ( 3a2 – 2a )]
Assume that 3a2 – 2a = x
Therefore,
5 – ( 3a2 – 2a )( 6 – 3a2 + 2a )
= 5 – x( 6 – x )
= 5 – 6x + x + x2
= 5( 1 – x ) – x( 1 – x )
= ( 5 – x )( 1 – x )
= ( x – 5 )( x – 1 )
= ( 3a2 – 2a – 5 )( 3a2 – 2a – 1 )
= ( 3a2 – 5a + 3a -5 )(3a2 – 3a + a – 1 )
= [ a( 3a – 5 ) + 1( 3a – 5)][3a( a – 1) + 1( a – 1)]
= ( 3a – 5 )( a + 1 )( 3a + 1 )( a – 1 )
Question 16
Factorise ……….
Answer
Question 17
Factories: (x2 – 3x)(x2 – 3x – 1) – 20.
Answer
(x2 – 3x)(x2 – 3x – 1) – 20
= (x2 – 3x)[(x2 – 3x) – 1] – 20
= a[a – 1] – 20 ….( Taking x2 – 3x = a )
= a2 – a – 20
= a2 – 5a + 4a – 20
= a(a – 5) + 4(a – 5)
= (a – 5)(a + 4)
= (x2 – 3x – 5)(x2 – 3x + 4)
Question 18.1
Find trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.
x2 – 3x – 54
Answer
Given expression : x2 – 3x – 54
Comparing with ax2 + bx + c, we get a = 1, b = -3, and c = – 54
∴ b2 – 4ac = (-3)2 – 4(1)(-54) = 9 + 216 = 225, which is a perfect square.
∴ x2 – 3x – 54 is factorisable.
Now, x2 – 3x – 54 = x2 – 9x + 6x – 54
= x( x – 9 ) + 6( x – 9 )
= ( x – 9 )( x + 6 )
Question 18.2
Find trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.
2x2 – 7x – 15
Answer
Given expression : 2x2 – 7x – 15
Comparing with ax2 + bx + c, we get a = 2, b = -7, and c = -15
∴ b2 – 4ac = (-7)2 – 4(2)(-15) = 49 + 120 = 169, which is a perfect square.
∴ 2x2 – 7x – 15 is factorisable.
Now, 2x2 – 7x – 15
= 2×2 – 10x + 3x – 15
= 2x( x – 5 ) + 3( x – 5 )
= ( 2x + 3 )( x – 5 )
Question 18.3
Find trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.
2x2 + 2x – 75
Answer
Given expression : 2x2 + 2x – 75
Comparing with ax2 + bx + c, we get a = 2, b = 2, and c = – 75
∴ b2 – 4ac = (2)2 – 4(2)(-75) = 4 + 600 = 604, which is not a perfect square.
∴ 2x2 + 2x – 75 is not factorisable.
Question 18.4
Find trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.
3x2 + 4x – 10
Answer
Given expression : 3x2 + 4x – 10
Comparing with ax2 + bx + c, we get a = 3, b = 4, and c = -10
∴ b2 – 4ac = (4)2 – 4(3)(-10) = 16 + 120 = 136, which is not a perfect square.
∴ 3x2 + 4x – 10 is not factorisable.
Question 18.5
Find trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.
x(2x – 1) – 1
Answer
Given expression : x(2x – 1) – 1
Now , x(2x – 1) – 1 = 2x2 – x – 1
Comparing with ax2 + bx + c, we get a = 2, b = – 1, and c = – 1
∴ b2 – 4ac = (- 1)2 – 4(2)(-1) = 1 + 8 = 9, which is a perfect square.
∴ 2x2 – x – 1 is factorisable.
Now, 2x2 – x – 1 = 2×2 – 2x + x – 1
= 2x( x – 1 ) + 1( x – 1 )
= ( 2x + 1 )( x – 1 )
Question 19.1
Factorise : 4√3x2 + 5x – 2√3
Answer
4√3x2 + 5x – 2√3
= 4√3x2 + 8x – 3x – 2√3
= 4x( √3x + 2 ) – √3( √3x + 2 )
= ( √3x + 2 )( 4x – √3 )
Question 19.2
Factorise : 7√2x2 – 10x – 4√2
Answer
7√2x2 – 10x – 4√2
= 7√2x2 – 14x + 4x – 4√2
= 7√2x( x – √2 ) + 4( x – √2 )
= ( x – √2 )( 7√2x + 4 )
Question 20
Give possible expressions for the length and the breadth of the rectangle whose area is 12x2 – 35x + 25
Answer
12x2 – 35x + 25
= 12x2 – 20x – 15x + 25
= 4x(3x – 5) – 5(3x – 5)
= (3x – 5)(4x – 5)
Thus,
Length = (3x – 5) and breadth = (4x – 5)
OR
Length = (4x – 5) and breadth = (3x – 5)
Factorisation Exercise-5 C, ICSE Class-9th Concise Selina Solutions
Question 1
Factorise : 25a2 – 9b2
Answer
25a2 – 9b2
= ( 5a )2 – ( 3b )2
= ( 5a – 3b )( 5a + 3b )
Question 2
Factorise : a2 – (2a + 3b)2
Answer
a2 – (2a + 3b)2
= a2 – ( 2a + 3b )2
= ( a – 2a – 3b )( a + 2a + 3b ) [ ∵ a2 – b2 = ( a + b )( a – b )]
= ( – a – 3b )( 3a + 3b )
= -3( a + 3b )( a + b )
Question 3
Factorise : a2 – 81 (b-c)2
Answer
a2 – 81 (b-c)2
= ( a )2 – [ 9( b – c ) ]2
= [ a – ( 9b – 9c )][ a + ( 9b – 9c )] [ ∵ a2 – b2 = ( a + b )( a – b )]
= ( a – 9b + 9c )( a + 9b – 9c )
Question 4
Factorise : 25(2a – b)2 – 81b2
Answer
25(2a – b)2 – 81b2
= [ 5( 2a – b )]2 – (9b)2
= [ 5( 2a – b ) – 9b ][ 5( 2a – b ) + 9b ]
[ ∵ a2 – b2 = ( a + b )( a – b )]
= [ 10a – 5b – 9b ][ 10a – 5b + 9b ]
= [ 10a – 14b ][ 10a + 4b ]
= 2 x ( 5a – 7b ) x 2 x ( 5a + 2b )
= 4( 5a – 7b )( 5a + 2b )
Question 5
Factorise : 50a3 – 2a
Answer
50a3 – 2a
= 2a( 25a2 – 1 )
= 2a[ (5a)2 – (1)2 ]
= 2a( 5a + 1 )( 5a – 1 ) [ ∵ a2 – b2 = ( a + b )( a – b )]
Question 6
Factorise : 4a2b – 9b3
Answer
4a2b – 9b3
= b( 4a2 – 9b2 )
= b[ (2a)2 – (3b)2 ]
= b( 2a – 3b )( 2a + 3b ) [ ∵ a2 – b2 = ( a + b )( a – b )]
Question 7
Factorise : 3a5 – 108a3
Answer
3a5 – 108a3
= 3a3( a2 – 36 )
= 3a3[( a )2 – ( 6 )2]
= 3a3( a – 6 )( a + 6 ) [ ∵ a2 – b2 = ( a + b )( a – b )]
Question 8
Factorise : 9(a – 2)2 – 16(a + 2)2
Answer
9(a – 2)2 – 16(a + 2)2
= [ 3( a – 2 )]2 – [4( a + 2 )]2
= [ 3( a – 2 ) – 4( a + 2 )][ 3( a – 2) + 4( a + 2 )]
[ ∵ a2 – b2 = ( a + b )( a – b )]
= [ 3a – 6 – 4a – 8 ][ 3a – 6 + 4a + 8 ]
= ( – a – 14 )( 7a + 2 )
= – ( a + 14 )( 7a + 2 )
Question 9
Factorise : a4 – 1
Answer
a4 – 1
= ( a2 )2 – ( 1 )2
= ( a2 + 1 )( a2 – 1 ) [ ∵ a2 – b2 = ( a + b )( a – b )]
= ( a2 + 1 )[ (a)2 – (1)2 ]
= ( a2 + 1 )( a + 1 )( a – 1 )
Question 10
Factorise : a3 + 2a2 – a – 2
Answer
a3 + 2a2 – a – 2
= a2( a + 2 ) – 1( a + 2 )
= ( a2 – 1 )( a + 2 )
= ( a + 1 )( a – 1 )( a + 2 ) [ ∵ a2 – b2 = ( a + b )( a – b )]
Question 11
Factorise : (a + b)3 – a – b
Answer
(a + b)3 – a – b
= ( a + b )3 – ( a + b )
= ( a + b )[ ( a + b )2 – 1 ]
= ( a + b )[ ( a + b )2 – (1)2 ]
= ( a + b )[( a + b ) + 1 ][( a + b ) – 1] [ ∵ a2 – b2 = ( a + b )( a – b )]
= ( a + b )( a + b + 1 )( a + b – 1 )
Question 12
Factorise : a (a – 1) – b (b – 1)
Answer
a (a – 1) – b (b – 1)
= a2 – a – b2 + b
= a2 – b2 – a + b
= ( a + b )( a – b ) – ( a – b ) [∵ a2 – b2 = ( a + b )( a – b )]
= ( a – b )[( a + b ) – 1]
= ( a – b )[ a + b – 1 ]
Question 13
Factorise : 4a2 – (4b2 + 4bc + c2)
Answer
4a2 – (4b2 + 4bc + c2)
= ( 2a )2 – ( 2b + c )2
= [ 2a – ( 2b + c )][ 2a + (2b + c )] [ ∵ a2 – b2 = ( a + b )( a – b )]
= [ 2a – 2b – c ][ 2a + 2b + c ]
Question 14
Factorise : 4a2 – 49b2 + 2a – 7b
Answer
4a2 – 49b2 + 2a – 7b
= [ ( 2a )2 – ( 7b )2] + [ 2a – 7b ]
= [ 2a – 7b ][ 2a + 7b ] + [ 2a – 7b ] [ ∵ a2 – b2 = ( a + b )( a – b ) ]
= [ 2a – 7b ][ 2a + 7b + 1 ]
Question 15
Factorise : 9a2 + 3a – 8b – 64b2
Answer
9a2 + 3a – 8b – 64b2
= 9a2 – 64b2 + 3a – 8b
= ( 3a )2 – ( 8b )2 + 3a – 8b
= ( 3a – 8b )( 3a + 8b ) + ( 3a – 8b )
[ ∵ a2 – b2 = ( a + b )( a – b )]
= ( 3a – 8b )( 3a + 8b + 1
Question 16
Factorise : 4a2 – 12a + 9 – 49b2
Answer
4a2 – 12a + 9 – 49b2
= ( 2a )2 – 12a + (3)2 – 49b2
= (2a – 3)2 – 49b2
= ( 2a – 3)2 – (7b)2
= ( 2a – 3 – 7b )( 2a – 3 + 7b ) [ ∵ a2 – b2 = ( a + b )( a – b )]
Question 17
Factorise : 4xy – x2 – 4y2 + z2
Answer
4xy – x2 – 4y2 + z2
= z2 – ( x2 + 4y2 – 4xy )
= z2 – ( x – 2y )2
= [ z – ( x – 2y )][ z + ( x – 2y )] [ ∵ a2 – b2 = ( a + b )( a – b )]
= [ z – x + 2y ][ z + x – 2y ]
Question 18
Factorise : a2 + b2 – c2 – d2 + 2ab – 2cd
Answer
a2 + b2 – c2 – d2 + 2ab – 2cd
= ( a2 + b2 + 2ab ) – ( c2 + d2 + 2cd )
= ( a + b )2 – ( c + d )2
= [( a + b ) – ( c + d )][( a + b ) + ( c + d )] [∵ a2 – b2 = ( a + b )( a – b )]
= ( a + b – c – d )( a + b + c + d )
Question 19
Factorise : 4x2 – 12ax – y2 – z2 – 2yz + 9a2
Answer
4x2 – 12ax – y2 – z2 – 2yz + 9a2
= 4x2 + 9a2 – 12ax – y2 – z2 – 2yz
= ( 2x )2 + ( 3a )2 – 12ax – ( y2 + z2 + 2yz )
= ( 2x – 3a )2 – ( y + z )2
= [( 2x – 3a ) – ( y + z )][( 2x – 3a ) + ( y + z )]
[ ∵ a2 – b2 = ( a + b )( a – b )]
= [ 2x – 3a – y – z ][ 2x – 3a + y + z ]
Question 20
Factorise : (a2 – 1) (b2 – 1) + 4ab
Answer
(a2 – 1) (b2 – 1) + 4ab
= a2b2 – a2 – b2 + 1 + 4ab
= a2b2 + 1 + 2ab – a2 – b2 + 2ab
= ( a2b2 + 1 + 2ab ) – ( a2 + b2 – 2ab )
= ( ab + 1)2 – ( a – b )2
= [( ab + 1 ) – ( a – b )][( ab + 1 ) + ( a – b )] [ ∵ a2 – b2 = ( a + b )( a – b )]
= [ ab + 1 – a + b ][ ab + 1 + a – b ]
Question 21
Factorise : x4 + x2 + 1
Answer
x4 + x2 + 1
= x4 + 2x2 + 1 – x2
= (x2)2 + 2x2 + (1)2 – x2
= ( x2 + 1 )2 – (x)2 [ ∵ a2 – b2 = ( a + b )( a – b )]
= ( x2 + 1 – x )( x2 + 1 + x )
Question 22
Factorise : (a2 + b2 – 4c2)2 – 4a2b2
Answer
(a2 + b2 – 4c2)2 – 4a2b2
= ( a2 + b2 – 4c2 )2 – ( 2ab )2
= ( a2 + b2 – 4c2 – 2ab )( a2 + b2 – 4c2 + 2ab ) [ ∵ a2 – b2 = ( a + b )( a – b )]
= ( a2 + b2 – 2ab – 4c2 )( a2 + b2 + 2ab – 4c2 )
= [ ( a – b )2 – ( 2c )2 ][ ( a + b )2 – ( 2c )2]
= ( a – b + 2c )( a – b – 2c )( a + b + 2c )( a + b – 2c )
Question 23
Factorise : (x2 + 4y2 – 9z2)2 – 16x2y2
Answer
(x2 + 4y2 – 9z2)2 – 16x2y2
= (x2 + 4y2 – 9z2)2 – ( 4xy )2
= ( x2 + 4y2 – 9z2 – 4xy )( x2 + 4y2 – 9z2 + 4xy ) [ ∵ a2 – b2 = ( a + b )( a – b )]
= ( x2 + 4y2 – 4xy – 9z2 )( x2 + 4y2 + 4xy – 9z2 )
= [( x – 2y )2 – (3z)2 ][ ( x + 2y )2 – (3z)2 ]
= [( x – 2y ) – 3z ][( x – 2y ) + 3z ][( x + 2y ) – 3z ][( x + 2y ) + 3z ]
= [ x – 2y – 3z ][ x – 2y + 3z ][ x + 2y – 3z ][ x + 2y + 3z ]
Question 24
Factorise : (a + b) 2 – a2 + b2
Answer
(a + b) 2 – a2 + b2
= a2 + 2ab + b2 – a2 + b2
= 2ab + 2b2
= 2b( a + b )
Question 25
Factorise : a2 – b2 – (a + b) 2
Answer
a2 – b2 – (a + b) 2
= a2 – b2 – ( a2 + 2ab + b2 )
= a2 – b2 – a2 – 2ab – b2
= – 2ab – 2b2
= – 2b( a + b )
Question 26
Factorize : 9a2 – (a2 – 4) 2
Answer
9a2 – (a2 – 4) 2
= ( 3a )2 – ( a2 – 4 )2
= [ 3a – ( a2 – 4 )][ 3a + ( a2 – 4 )]
= [ 3a – a2 – 4 ][ 3a + a2 – 4 ]
= [ -a2 + 3a – 4 ][ a2 + 3a – 4 ]
= [ -a2 + 4a – a – 4 ][ a2 + 4a – a – 4 ]
= [ a( -a + 4 ) + 1( -a + 4 )][ a( a + 4 ) – 1( a + 4 )]
= [( a + 1 )( 4 – a )][( a + 4 )( a – 1 )]
= ( a + 1 )( 4 – a )( a + 4 )( a – 1 )
Question 27
Factorise :
Answer
Question 28
Factorise :
Answer
Question 29
Factorise : 4x4 – x2 – 12x – 36
Answer
Factorise : 4x4 – x2 – 12x – 36
= 4x4 – ( x2 + 12x + 36 )
= ( 2x2)2 – ( x2 + 2 x x x 6 + 62 )
= ( 2x2)2 – ( x + 6 )2
= ( 2x2 + x + 6 )( 2x2 – x – 6 )
= ( 2x2 + x + 6 )( 2x2 – 4x + 3x – 6 )
= ( 2x2 + x + 6 )[ 2x( x – 2 ) + 3( x – 2 )]
= ( 2x2 + x + 6 )[ ( x – 2)( 2x + 3 )]
= ( 2x2 + x + 6 )( x – 2 )( 2x + 3 )
Question 30
Factorise : a2 ( b + c) – (b + c)3
Answer
a2 ( b + c) – (b + c)3
= ( b + c )[ a2 – ( b + c )2 ]
= ( b + c )[( a + b + c )( a – b – c )]
= ( b + c )( a + b + c )( a – b – c )
Concise Maths Exercise – 5(D), Factorisation ICSE Class-9th Selina Solutions
Question 1
Factorise : a3 – 27
Answer
a3 – 27
= ( a )3 – ( 3 )3
= ( a – 3 )[ (a)2 + a x 3 + (3)2 ] [ ∵ a3 – b3 = ( a – b )( a2 + ab + b2 )]
= ( a – 3 )[ a2 + 3a + 9 ]
Question 2
Factorise : 1 – 8a3
Answer
1 – 8a3
= (1)3 – (2a)3
= ( 1 – 2a )[ (1)2 + 1 x 2a + (2a)2 ]
[ ∵ a3 – b3 = ( a – b )( a2 + ab + b2 )]
= ( 1 – 2a )[ 1 + 2a + 4a2 ]
Question 3
Factorise : 64 – a3b3
Answer
64 – a3b3
= (4)3 – (ab)3
= ( 4 – ab )[(4)2 + 4 x ab + (ab)2 ] [ ∵ a3 – b3 = ( a – b )( a2 + ab + b2 )]
= ( 4 – ab )( 16 + 4ab + a2b2 )
Question 4
Factorise : a6 + 27b3
Answer
a6 + 27b3
= ( a2 )3 + ( 3b )3
= ( a2 + 3b )[ (a2)2 – a2 x 3b + (3b)2 ] [ ∵ a3 + b3 = ( a + b )( a2 – ab + b2 )]
= ( a2 + 3b )[ a4 – 3a2b + 9b2 ]
Question 5
Factorise : 3x7y – 81x4y4
Answer
3x7y – 81x4y4
= 3xy( x6 – 27x3y3 )
= 3xy[ (x2)3 – ( 3xy )3 ]
= 3xy( x2 – 3xy )[ (x2)2 + x2 x 3xy + (3xy)2 ] [ ∵ a3 – b3 = ( a -b )( a2 + ab + b2 )]
= 3xy( x2 – 3xy )[ x4 + 3x3y + 9x2y2 ]
= 3xy [ x( x + 3y) x2( x2 + 3xy + 9y2 ) ]
= 3x4y( x – 3y )( x2 + 3xy + 9y2 )
Question 6
Factorise : a3…….
Answer
Question 7
Factorise : a3 + 0.064
Answer
a3 + 0.064
= (a)3 + (0.4)3
= ( a + 0.4 )[ (a)2 – a x 0.4 + (0.4)2 ] [ ∵ a3 + b3 = ( a + b )( a2 – ab + b2 ) ]
= ( a + 0.4 )( a2 – 0.4a + 0.16 )
Question 8
Factorise : a4 – 343a
Answer
a4 – 343a
= a( a3 – 73 )
= a( a – 7 )[(a)2 + a x 7 + (7)2 ] [ ∵ a3 – b3 = ( a – b )( a2 + ab + b2 )]
= a( a – 7 )( a2 + 7a + 49 )
Question 9
Factorise: (x – y)3 – 8x3
Answer
( x – y )3 – 8x3
= ( x – y )3 – ( 2x )3
= ( x – y – 2x )[ (x – y)2 + 2x(x – y) + (2x)2 ]
[ Using identity (a3 – b3) = (a – b)(a2 + ab + b2) ]
= (- x – y ) [ x2 + y2 – 2xy + 2x2 – 2xy + 4x2 ]
=- ( x + y ) [ 7x2 – 4xy + y2 ]
Question 10
Factorise : ……
Answer
Question 11
Factorise : a6 – b6
Answer
We know that,
a3 + b3 = ( a + b )( a2 – ab + b2 ) ….(1)
a3 – b3 = ( a – b )( a2 + ab + b2 ) ….(2)
a6 – b6
= ( a3)2 – (b3)2
= ( a3 + b3 )( a3 – b3 )
= ( a + b )( a2 – ab + b2 )( a – b )( a2 + ab + b2 )
[ From(1) and (2) ]
= ( a + b )( a – b )( a2 – ab + b2 )( a2 + ab + b2 )
Question 12
Factorise : a6 – 7a3 – 8
Answer
We know that,
a3 + b3 = ( a + b )( a2 – ab + b2 ) ….(1)
a3 – b3 = ( a – b )( a2 + ab + b2 ) ….(2)
a6 – 7a3 – 8
= a6 – 8a3 + a3 – 8
= a3( a3 – 8) + 1( a3 – 8 )
= ( a3 + 1 )( a3 – 8 )
= ( a3 + 13 )( a3 – 23 )
= ( a + 1 )( a2 – a + 1 )( a – 2 )( a2 + 2a + 4 )
[ From(1) and (2) ]
= ( a + 1 )( a – 2)( a2 – a + 1 )( a2 + 2a + 4 )
Question 13
Factorise : a3 – 27b3 + 2a2b – 6ab2
Answer
a3 – 27b3 + 2a2b – 6ab2
We know that,
a3 – b3 = ( a – b )( a2 + ab + b2 ) ….(1)
a3 – 27b3 + 2a2b – 6ab2
= (a)3 – (3b)3 + 2ab( a – 3b )
= ( a – 3b )[ a2 + a x 3b + (3b)2 ] + 2ab( a – 3b ) [From(1)]
= ( a – 3b )[ a2 + 3ab + 9b2 ] + 2ab( a – 3b )
= ( a – 3b )[ a2 + 3ab + 9b2 + 2ab ]
= ( a – 3b )[ a2 + 5ab + 9b2 ]
Factorise : 8a3 – b3 – 4ax + 2bx
Answer
We know that,
a3 – b3 = ( a – b )( a2 + ab + b2 ) …..(1)
8a3 – b3 – 4ax + 2bx
= [ (2a)3 – (b)3 ] – 2 x ( 2a – b )
= ( 2a – b )[ (2a)2 + 2a x b + (b)2 ] – 2 x ( 2a – b ) [ From(1) ]
= ( 2a – b )[ 4a2 + 2ab + b2 ] – 2 x ( 2a – b )
= ( 2a – b )[ 4a2 + 2ab + b2 – 2x ]
Question 15
Factorise : a – b – a3 + b3
Answer
we know that,
a3 – b3 = ( a – b )( a2 + ab + b2 ) ….(1)
a – b – a3 + b3
= a – b – ( a3 – b3 )
= ( a – b ) – ( a – b )[ a2 + ab + b2 ] [ From (1) ]
= ( a – b )[ 1 – a2 – ab – b2 ]
Question 16
Factorise : 2x3 + 54y3 – 4x – 12y
Answer
2x3 + 54y3 – 4x – 12y
= 2 ( x3 + 27y3 – 2x – 6y )
= 2 [ { (x)3+ (3y)3} – 2(x + 3y) ]
[ Using identity (a3 + b3) = (a + b)(a2 – ab + b2) ]
= 2[ {(x + 3y)(x2 – 3xy + 9y2)} – 2(x + 3y) ]
= 2(x + 3y)(x2 – 3xy + 9y2 – 2)
Question 17
Factorise : 1029 – 3x3
Answer
1029 – 3x3
= 3( 343 – x3 )
= 3( 73 – x3 )
= 3( 7 – x )( 72 + 7x + x2 )
= 3( 7 – x )( 49 + 7x + x2 )
Question 18.1
Show that : 133 – 53 is divisible by 8
Answer
( 133 – 53 )
[ Using identity (a3 – b3) = (a – b)(a2 + ab + b2)]
= ( 13 – 5 )( 132 + 13 × 5 + 52 )
= 8( 169 + 65 + 25 )
Therefore, the number is divisible by 8.
Question 18.2
Show that : 353 + 273 is divisible by 62
Answer
(353 + 273)
[Using identity (a3 + b3)=(a + b)(a2 – ab + b2)]
= ( 35 + 27 )( 352 + 35× 27 + 272 )
= 62 × ( 352 + 35 × 27 + 272 )
Therefore, the number is divisible by 62.
Question 19
Evaluate :………………………..
Answer
Let a = 5.67 and b = 4.33
Then,
Exercise – 5(E)
(Factorisation ICSE Class-9th Concise Mathematics Selina Solutions Chapter-5.)
Question 1
Factorise :
Answer
Question 2
Factorise :
Answer
Question 3
Factorise :
Answer
Question 4
Factorise :
Answer
Question 5
Factorise : 4x4 + 9y4 + 11x2y2
Answer
4x4 + 9y4 + 11x2y2
= (2x2)2 + (3y2)2 + 12x2y2 – x2y2
= (2x2 + 3y2)2 – x2y2
= (2x2 + 3y2)2 – (xy)2
= ( 2x2 + 3y2 – xy )( 2x2 + 3y2 + xy) [ ∵ a2 – b2 = ( a + b )( a – b )]
Question 6
Factorise :
Answer
Question 7
Factorise : a – b – 4a2 + 4b2
Answer
a – b – 4a2 + 4b2
= ( a – b ) – 4( a2 – b2 )
= ( a – b ) – 4( a – b )( a + b ) [ ∵ a2 – b2 = ( a + b )( a – b )]
= ( a – b )[ 1 – 4( a + b )]
= ( a – b )[ 1 – 4a – 4b ]
Question 8
Factorise : (2a – 3)2 – 2 (2a – 3) (a – 1) + (a – 1)2
Answer
(2a – 3)2 – 2 (2a – 3) (a – 1) + (a – 1)2
= [( 2a – 3 ) – ( a – 1 )]2
= [ 2a – 3 – a + 1 ]2
= ( a – 2 )2
Question 9
Factorise : (a2 – 3a) (a2 + 3a + 7) + 10
Answer
(a2 – 3a) (a2 + 3a + 7) + 10
Let us assume , a2 – 3a = x
Then the given expression is,
( a2 – 3a )( a2 – 3a + 7 ) + 10
= x( x + 7 ) + 10
= x2 + 7x + 10
= x2 + 5x + 2x + 10
= x( x + 5 ) + 2 ( x + 5 )
= ( x + 5 )( x + 2 )
= ( a2 – 3a + 5 )( a2 – 3a + 2 ) [resubstituting the value of x ]
= ( a2 – 3a + 5 )( a2 – 2a – a + 2 )
= ( a2 – 3a + 5 )( a( a – 2) – 1(a – 2))
= ( a2 – 3a + 5 )[( a – 1 )( a – 2 )]
Question 10
Factorise : (a2 – a) (4a2 – 4a – 5) – 6
Answer
Let us assume, a2 – a = x
Then the given expression is,
(a2 – a) (4a2 – 4a – 5) – 6
= x( 4x – 5 ) – 6
= 4x2 – 5x – 6
= 4x2 – 8x + 3x – 6
= 4x( x – 2 ) + 3( x – 2 )
= ( 4x + 3 )( x – 2 )
= [ 4( a2 – a ) + 3 ]( a2 – a – 2 ) [ resubstitute the value of x ]
= [ 4a2 – 4a + 3 ]( a2 – a – 2 )
= [ 4a2 – 4a + 3 ]( a2 – 2a + a – 2 )
= [ 4a2 – 4a + 3 ][ a( a – 2 ) + 1( a – 2 )]
= [ 4a2 – 4a + 3 ]( a – 2 )( a + 1 )
Question 11
Factorise : x4 + y4 – 3x2y2
Answer
x4 + y4 – 3x2y2
= x4 + y4 – 2x2y2 – x2y2
= (x2)2 + (y2)2 – 2x2y2 – x2y2
= ( x2 – y2 )2 – (xy)2
= ( x2 – y2 – xy )( x2 – y2 + xy ) [ ∵ a2 – b2 = ( a + b )( a – b )]
Question 12
Factorise : 5a2 – b2 – 4ab + 7a – 7b
Answer
5a2 – b2 – 4ab + 7a – 7b
= 4a2 + a2 – b2 – 4ab + 7a – 7b
= a2 – b2 + 4a2 – 4ab + 7a – 7b
= ( a2 – b2 ) + 4a( a – b ) + 7( a – b )
= ( a – b )( a + b ) + 4a( a – b ) + 7( a – b ) [ ∵ a2 – b2 = ( a + b )( a – b ) ]
= ( a – b )[ ( a + b ) + 4a + 7 ]
= ( a – b )[ ( a + b ) + 4a + 7 ]
= ( a – b )[ 5a + b + 7 ]
Question 13
Factorise : 12(3x – 2y)2 – 3x + 2y – 1
Answer
12(3x – 2y)2 – 3x + 2y – 1 = 12( 3x – 2y )2 – ( 3x – 2y ) – 1
Let us assume that 3x – 2y = a
Then the given expression is
12(3x – 2y)2 – 3x + 2y – 1
= 12a2 – 3a – 1
= 12a2 – 4a + 3a – 1
= 4a( 3a – 1 ) + 1( 3a – 1 )
= ( 4a + 1 )( 3a – 1 )
= [ 4( 3x – 2y ) + 1 ][ 3( 3x – 2y ) – 1 ] [ resubstitute the value of a ]
= ( 12x – 8y + 1 )( 9x – 6y – 1)
Factorise : 4(2x – 3y)2 – 8x+12y – 3
Answer
4(2x – 3y)2 – 8x+12y – 3
= 4(2x – 3y)2 – 4(2x – 3y) – 3
Let us assume that 2x – 3y = a
Then the given expression is
4(2x – 3y)2 – 8x+12y – 3
= 4a2 – 4a – 3
= 4a2 – 6a + 2a – 3
= 2a( 2a – 3 ) + 1( 2a – 3)
= ( 2a – 3 )( 2a + 1 )
= [ 2( 2x – 3y ) – 3 ][ 2( 2x – 3y ) + 1 ]
= ( 4x – 6y – 3 )( 4x – 6y + 1 )
Question 15
Factorise : 3 – 5x + 5y – 12(x – y)2
Answer
3 – 5x + 5y – 12(x – y)2 = 3 – 5( x – y ) – 12(x – y)2
Let us assume that x – y = a
Then the given expression is
3 – 5x + 5y – 12(x – y)2
= 3 – 5a – 12a2
= 3 – 9a + 4a – 12a2
= 3( 1 – 3a ) + 4a( 1 – 3a )
= ( 3 + 4a )( 1 – 3a ) [ resubstitute the value of a ]
= [ 3 + 4( x – y )][ 1 – 3( x – y )]
= ( 3 + 4x – 4y )( 1 – 3x + 3y )
Question 16
Factorise : 9x 2 + 3x – 8y – 64y2
Answer
9x 2 + 3x – 8y – 64y2
= 9x2 – 64y2 + 3x – 8y
= [ (3x)2 – (8y)2 ] + ( 3x – 8y )
= [( 3x + 8y )( 3x – 8y )] + ( 3x – 8y )
= ( 3x – 8y )( 3x + 8y + 1 )
Question 17
Factorise : 2√3x2 + x – 5√3
Answer
2√3x2 + x – 5√3
= 2√3x2 + 6x – 5x – 5√3
= 2√3x( x + √3 ) – 5( x + √3 )
= ( 2√3x – 5 )( x + √3 )
Question 18
Factorise :
Answer
Question 19
Factorise : 2(ab + cd) – a2 – b2 + c2 + d2
Answer
2(ab + cd) – a2 – b2 + c2 + d2
= 2ab + 2cd – a2 – b2 + c2 + d2
= c2 + d2 + 2cd – a2 – b2 + 2ab
= ( c2 + d2 + 2cd ) – ( a2 + b2 – 2ab )
= ( c + d )2 – ( a – b )2
= ( c + d + a – b )( c + d – a + b )
Question 20.1
Find the value of : ( 987 )2 – (13)2
Answer
( 987 )2 – (13)2
= ( 987 + 13 )( 987 – 13)
= 1000 x 974
= 974000
Question 20.2
Find the value of : ( 67.8 )2 – ( 32.2 )2
Answer
( 67.8 )2 – ( 32.2 )2
= ( 67.8 + 32.2 )( 67.8 – 32.2 )
= 100 x 35.6
= 3560
Question 20.3
Find the value of :………..
Answer
Question 20.4
Find the value of : …………………
Answer
— End of Factorisation ICSE Class-9th Concise Solutions :–
Return to – Concise Selina Maths Solutions for ICSE Class -9
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