RD Sharma Solutions for Class 9 Maths Chapter 5–Factorization of Algebraic Expressions
RD Sharma Solutions for Class 9 Maths Chapter 5–Factorization of Algebraic Expressions

Class 9: Maths Chapter 5 solutions. Complete Class 9 Maths Chapter 5 Notes.

RD Sharma Solutions for Class 9 Maths Chapter 5–Factorization of Algebraic Expressions

RD Sharma 9th Maths Chapter 5, Class 9 Maths Chapter 5 solutions

Exercise 5.1 Page No: 5.9

Question 1: Factorize x3 + x – 3x2 – 3

Solution:

x3 + x – 3x2 – 3

Here x is common factor in x3 + x and – 3 is common factor in – 3x2 – 3

x3 – 3x2 + x – 3

x2 (x – 3) + 1(x – 3)

Taking ( x – 3) common

(x – 3) (x2 + 1)

Therefore x3 + x – 3x2 – 3 = (x – 3) (x2 + 1)

Question 2: Factorize a(a + b)3 – 3a2b(a + b)

Solution:

a(a + b)3 – 3a2b(a + b)

Taking a (a + b) as common factor

= a(a + b) {(a + b)2 – 3ab}

= a(a + b) {a2 + b2 + 2ab – 3ab}

= a(a + b) (a2 + b2 – ab)

Question 3: Factorize x(x3 – y3) + 3xy(x – y)

Solution:

x(x3 – y3) + 3xy(x – y)

= x(x – y) (x2 + xy + y2) + 3xy(x – y)

Taking x(x – y) as a common factor

= x(x – y) (x2 + xy + y2 + 3y)

= x(x – y) (x2 + xy + y2 + 3y)

Question 4: Factorize 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)

Question 5: Factorize x2 + y – xy – x

Solution:

x2 + y – xy – x

= x2 – x – xy + y

= x(x- 1) – y(x – 1)

= (x – 1) (x – y)

Question 6: Factorize x3 – 2x2y + 3xy2 – 6y3

Solution:

x3 – 2x2y + 3xy2 – 6y3

= x2(x – 2y) + 3y2(x – 2y)

= (x – 2y) (x2 + 3y2)

Question 7: Factorize 6ab – b2 + 12ac – 2bc

Solution:

6ab – b2 + 12ac – 2bc

= 6ab + 12ac – b2 – 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 (x2 + 1/x2) – 4(x + 1/x) + 6

Solution:

(x2 + 1/x2) – 4(x + 1/x) + 6

= x2 + 1/x2 – 4x – 4/x + 4 + 2

= x2 + 1/x2 + 4 + 2 – 4/x – 4x

= (x2) + (1/x) 2 + ( -2 )2 + 2x(1/x) + 2(1/x)(-2) + 2(-2)x

As we know, x2 + y2 + z2 + 2xy + 2yz + 2zx = (x+y+z) 2

So, we can write;

= (x + 1/x + (-2 )) 2

or (x + 1/x – 2) 2

Therefore, x2 + 1/x2) – 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) (x2 – 4x + 4)

= (x – 2) [x2 – 2 (x)(2) + (2) 2]

= (x – 2) (x – 2) 2

= (x – 2)3

Question 10: Factorize ( x + 2 ) ( x2 + 25 ) – 10x2 – 20x

Solution :

( x + 2) ( x2 + 25) – 10x ( x + 2 )

Take ( x + 2 ) as common factor;

= ( x + 2 )( x2 + 25 – 10x)

=( x + 2 ) ( x2 – 10x + 25)

Expanding the middle term of ( x2 – 10x + 25 )

=( x + 2 ) ( x2 – 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) ( x2 + 25) – 10x ( x + 2 ) = ( x + 2 )( x – 5 )2

Question 11: Factorize 2a2 + 2√6 ab + 3b2

Solution:

2a2 + 2√6 ab + 3b2

Above expression can be written as ( √2a )2 + 2 × √2a × √3b + ( √3b)2

As we know, ( p + q ) 2 = p2 + q2 + 2pq

Here p = √2a and q = √3b

= (√2a + √3b )2

Therefore, 2a2 + 2√6 ab + 3b2 = (√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 p2 + q2 + 2pq = (p + q) 2}

Here p = a – b + c and q = b – c + a

= [a – b + c + b- c + a]2

= (2a)2

= 4a2

Question 13: Factorize a2 + b2 + 2( ab+bc+ca )

Solution:

a2 + b2 + 2ab + 2bc + 2ca

As we know, p2 + q2 + 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, a2 + b2 + 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

= 4p2 – 12pq + 9q2

Expanding the middle term, -12 = -6 -6 also 4× 9=-6 × -6

= 4p2 – 6pq – 6pq + 9q2

=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 a2 – b2 + 2bc – c2

Solution :

a2 – b2 + 2bc – c2

As we know, ( a-b)2 = a2 + b2 – 2ab

= a2 – ( b – c) 2

Also we know, a2 – b2 = ( a+b)( a-b)

= ( a + b – c )( a – ( b – c ))

= ( a + b – c )( a – b + c )

Therefore, a2 – b2 + 2bc – c2 =( a + b – c )( a – b + c )

Question 16: Factorize a2 + 2ab + b2 – c2

Solution:

a2 + 2ab + b2 – c2

= (a2 + 2ab + b2) – c2

= (a + b)2 – (c) 2

We know, a2 – b2 = (a + b) (a – b)

= (a + b + c) (a + b – c)

Therefore a2 + 2ab + b2 – c2 = (a + b + c) (a + b – c)

Exercise 5.2 Page No: 5.13

Factorize each of the following expressions:

Question 1: p3 + 27

Solution:

p3 + 27

= p3 + 33[using a3 + b3 = (a + b)(a2 –ab + b2)]

= (p + 3)(p² – 3p – 9)

Therefore, p3 + 27 = (p + 3)(p² – 3p – 9)

Question 2: y3 + 125

Solution:

y3 + 125

= y3 + 53[using a3 + b3 = (a + b)(a2 –ab + b2)]

= (y+5)(y2 − 5y + 52)

= (y + 5)(y2 − 5y + 25)

Therefore, y3 + 125 = (y + 5)(y2 − 5y + 25)

Question 3: 1 – 27a3

Solution:

= (1)3 −(3a) 3[using a3 – b3 = (a – b)(a2 + ab + b2)]

= (1− 3a)(12 + 1×3a + (3a) 2)

= (1−3a)(1 + 3a + 9a2)

Therefore, 1−27a3 = (1−3a)(1 + 3a+ 9a2)

Question 4: 8x3y3 + 27a3

Solution:

8x3y3 + 27a3

= (2xy) 3 + (3a) 3[using a3 + b3 = (a + b)(a2 –ab + b2)]

= (2xy +3a)((2xy)2−2xy×3a+(3a) 2)

= (2xy+3a)(4x2y2 −6xya + 9a2)

Question 5: 64a3 − b3

Solution:

64a3 − b3

= (4a)3−b3[using a3 – b3 = (a – b)(a2 + ab + b2)]

= (4a−b)((4a)2 + 4a×b + b2)

=(4a−b)(16a2 +4ab+b2)

Question 6: x3 / 216 – 8y3

Solution:

x3 / 216 – 8y3

Question 7: 10x4 y – 10xy4

Solution:

10x4 y – 10xy4

= 10xy(x3 − y3)[using a3 – b3 = (a – b)(a2 + ab + b2)]

= 10xy (x−y)(x2 + xy + y2)

Therefore, 10x4 y – 10xy4 = 10xy (x−y)(x2 + xy + y2)

Question 8: 54x6 y + 2x3y4

Solution:

54x6 y + 2x3y4

= 2x3y(27x3 +y3)

= 2x3y((3x) 3 + y3)[using a3 + b3 = (a + b)(a2 – ab + b2)]

= 2x3y {(3x+y) ((3x)2−3xy+y2)}

=2x3y(3x+y)(9x2 − 3xy + y2)

Question 9: 32a3 + 108b3

Solution:

32a3 + 108b3

= 4(8a3 + 27b3)

= 4((2a) 3+(3b) 3)[using a3 + b3 = (a + b)(a2 – ab + b2)]

= 4[(2a+3b)((2a)2−2a×3b+(3b) 2)]

= 4(2a+3b)(4a2 − 6ab + 9b2)

Question 10: (a−2b)3 − 512b3

Solution:

(a−2b)3 − 512b3

= (a−2b)3 −(8b) 3[using a3 – b3 = (a – b)(a2 + ab + b2)]

= (a −2b−8b) {(a−2b)2 + (a−2b)8b + (8b) 2}

=(a −10b)(a2 + 4b2 − 4ab + 8ab − 16b2 + 64b2)

=(a−10b)(a2 + 52b2 + 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 p3 – q3 = (p – q)(p2 + pq + q2)]

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)(a2+b2+2ab+(a+b)(2a−2b)+(2a−2b) 2)

=(a+b−2a+2b)(a2+b2+2ab+2a2−2ab+2ab−2b2+(2a−2b) 2)

=(3b−a)(3a2+2ab−b2+(2a−2b) 2)

=(3b−a)(3a2+2ab−b2+4a2+4b2−8ab)

=(3b−a)(3a2+4a2−b2+4b2−8ab+2ab)

=(3b−a)(7a2+3b2−6ab)

Question 12: (x+2)3 + (x−2) 3

Solution:

(x+2)3 + (x−2) 3[using p3 + q3 = (p + q)(p2 – pq + q2)]

Here p = x + 2 and q = x – 2

= (x+2+x−2)((x+2)2−(x+2)(x−2)+(x−2) 2)

=2x(x2 +4x+4−(x+2)(x−2)+x2−4x+4)[ Using : (a+b)(a−b) = a2−b2 ]

= 2x(2x2 + 8 − (x2 − 22))

= 2x(2x2 +8 − x2 + 4)

= 2x(x2 + 12)

Exercise 5.3 Page No: 5.17

Question 1: Factorize 64a3 + 125b3 + 240a2b + 300ab2

Solution:

64a3 + 125b3 + 240a2b + 300ab2

= (4a)3 + (5b) 3 + 3(4a)2(5b) + 3(4a)(5b)2 , which is similar to a3 + b3 + 3a2b + 3ab2

We know that, a3 + b3 + 3a2b + 3ab2 = (a+b)3]

= (4a+5b)3

Question 2: Factorize 125x3 – 27y3 – 225x2y + 135xy2

Solution:

125x3 – 27y3 – 225x2y + 135xy2

Above expression can be written as (5x)3−(3y) 3−3(5x)2(3y) + 3(5x)(3y)2

Using: a3 − b3 − 3a2b + 3ab2 = (a−b)3

= (5x − 3y)3

Question 3: Factorize 8/27 x3 + 1 + 4/3 x2 + 2x

Solution:

8/27 x3 + 1 + 4/3 x2 + 2x

Question 4: Factorize 8x3 + 27y3 + 36x2y + 54xy2

Solution:

8x3 + 27y3 + 36x2y + 54xy2

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, 8x3 + 27y3 + 36x2y + 54xy2 = (2x+3y)3

Question 5: Factorize a3 − 3a2b + 3ab2 − b3 + 8

Solution:

a3 − 3a2b + 3ab2 − b3 + 8

Using: a3 − b3 − 3a2b + 3ab2 = (a−b)3

= (a−b)3 + 23

Again , Using: a3 + b3 =(a + b)(a2 – ab + b2)]

=(a−b+2)((a−b)2−(a−b) × 2 + 22)

=(a−b+2)(a2+b2−2ab−2(a−b)+4)

=(a−b+2)(a2+b2−2ab−2a+2b+4)

a3 − 3a2b + 3ab2 − b3 + 8 =(a−b+2)(a2+b2−2ab−2a+2b+4)

Exercise 5.4 Page No: 5.22

Factorize each of the following expressions:

Question 1: a3 + 8b3 + 64c3 − 24abc

Solution:

a3 + 8b3 + 64c3 − 24abc

= (a)3 + (2b) 3 + (4c) 3− 3×a×2b×4c[Using a3+b3+c3−3abc = (a+b+c)(a2+b2+c2−ab−bc−ca)]

= (a+2b+4c)(a2+(2b)2 + (4c)2−a×2b−2b×4c−4c×a)

= (a+2b+4c)(a2 +4b2 +16c2 −2ab−8bc−4ac)

Therefore, a3 + 8b3 + 64c3 − 24abc = (a+2b+4c)(a2 +4b2 +16c2 −2ab−8bc−4ac)

Question 2: x 3 − 8y 3+ 27z3 + 18xyz

Solution:

= x3 − (2y) 3 + (3z) 3 − 3×x×(−2y)(3z)

= (x + (−2y) + 3z) (x2 + (−2y)2 + (3z) 2 −x(−2y)−(−2y)(3z)−3z(x))[using a3+b3+c3−3abc = (a+b+c)(a2+b2+c2−ab−bc−ca)]

=(x −2y + 3z)(x2 + 4y2 + 9 z2 + 2xy + 6yz − 3zx)

Question 3: 27x 3 − y 3– z3 – 9xyz

Solution:

27x 3 − y 3– z3 – 9xyz

= (3x) 3 − y 3– z3 – 3(3xyz)[Using a3 + b3 + c3 −3abc = (a + b + c)(a2+b2+c2−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){ 9x2 + y2 + z2 + 3xy – yz + 3xz)}

Question 4: 1/27 x3 − y3 + 125z3 + 5xyz

Solution:

1/27 x3 − y3 + 125z3 + 5xyz

= (x/3)3+(−y)3 +(5z)3 – 3 x/3 (−y)(5z)[Using a3 + b3 + c3 −3abc = (a + b + c)(a2+b2+c2−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 + y2 + 25z2 + xy/3 + 5yz – 5zx/3)

Question 5: 8x3 + 27y3 − 216z3 + 108xyz

Solution:

8x3 + 27y3 − 216z3 + 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) {4x2 +9y2 +36z2 −6xy + 18yz + 12zx}

Question 6: 125 + 8x3 − 27y3 + 90xy

Solution:

125 + 8x3 − 27y3 + 90xy

= (5)3 + (2x) 3 +(−3y) 3 −3×5×2x×(−3y)

= (5+2x+(−3y)) (52 +(2x) 2 +(−3y) 2 −5(2x)−2x(−3y)−(−3y)5)

= (5+2x−3y)(25+4x2 +9y2 −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, a3 + b3 + c3 −3abc = (a + b + c)(a2+b2+c2−ab−bc−ca)

⇒ a3 + b3 + c3 −3abc = 0

or a3 + b3 + c3 =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, a3 + b3 + c3 −3abc = (a + b + c)(a2+b2+c2−ab−bc−ca)

⇒ a3 + b3 + c3 −3abc = 0

(2x−3y)3 + (4z−2x) 3 + (3y−4z) 3 = 3(2x−3y)(4z−2x)(3y−4z)

Exercise VSAQs Page No: 5.24

Question 1: Factorize x4 + x2 + 25

Solution:

x4 + x2 + 25

= (x2) 2 + 52 + x2[using a2 + b2 = (a + b) 2 – 2ab ]

= (x2 +5) 2 −2(x2 ) (5) + x2

=(x2 +5) 2 −10x2 + x2

=(x2 + 5) 2 − 9x2

=(x2 + 5) 2 − (3x) 2[using a2 – b2 = (a + b)(a – b ]

= (x 2 + 3x + 5)(x2 − 3x + 5)

Question 2: Factorize x2 – 1 – 2a – a2

Solution:

x2 – 1 – 2a – a2

x2 – (1 + 2a + a2 )

x2 – (a + 1)2

(x – (a + 1)(x + (a + 1)

(x – a – 1)(x + a + 1)[using a2 – b2 = (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 a3 + b3 + c3.

Solution:

We know, a3 + b3 + c3 – 3abc = (a + b +c ) (a2 + b2 + c2 – ab – bc − ca)

Put a + b + c =0

This implies

a3 + b3 + c3 = 3abc

Question 4: If a2 + b2 + c2 = 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 a2 + b2 + c2 .

Solution:

We know, (a+b+c)² = a² + b² + c² + 2(ab + bc + ca)

92 = a² + b² + c² + 2(40)

81 = a² + b² + c² + 80

⇒ a² + b² + c² = 1

RD Sharma Solutions for Class 9 Maths Chapter 5: Download PDF

RD Sharma Solutions for Class 9 Maths Chapter 5–Factorization of Algebraic Expressions

Download PDF: RD Sharma Solutions for Class 9 Maths Chapter 5–Factorization of Algebraic Expressions PDF

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About RD Sharma

RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.

R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.

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