RD Sharma Solutions Class 9 Maths Chapter 5 Factorisation of Algebraic Expressions

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 Class 9 Solutions Maths
RD Sharma Class 9 Solutions Maths Chapter 1	RD Sharma Class 9 Solutions Maths Chapter 2
RD Sharma Class 9 Solutions Maths Chapter 3	RD Sharma Class 9 Solutions Maths Chapter 4
RD Sharma Solutions Class 9 Maths Chapter 5	RD Sharma Class 9 Solutions Maths Chapter 6
RD Sharma Class 9 Solutions Maths Chapter 7	RD Sharma Class 9 Solutions Maths Chapter 8
RD Sharma Class 9 Solutions Maths Chapter 9	RD Sharma Class 9 Solutions Maths Chapter 10
RD Sharma Class 9 Solutions Maths Chapter 11	RD Sharma Class 9 Solutions Maths Chapter 12
RD Sharma Class 9 Solutions Maths Chapter 13	RD Sharma Class 9 Solutions Maths Chapter 14
RD Sharma Class 9 Solutions Maths Chapter 15	RD Sharma Class 9 Solutions Maths Chapter 16
RD Sharma Class 9 Solutions Maths Chapter 17	RD Sharma Class 9 Solutions Maths Chapter 18
RD Sharma Class 9 Solutions Maths Chapter 19	RD Sharma Class 9 Solutions Maths Chapter 20
RD Sharma Class 9 Solutions Maths Chapter 21	RD Sharma Class 9 Solutions Maths Chapter 22
RD Sharma Class 9 Solutions Maths Chapter 23	RD Sharma Class 9 Solutions Maths Chapter 24

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

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