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

RD Sharma Solutions for Class 9 Maths Chapter 5 – Free PDF Download

RD Sharma Solutions for Class 9 Maths Chapter 5 – Factorization of Algebraic Expressions are given here in detail. These study materials include five exercises, and all the questions are solved by BYJU’S subject experts. RD Sharma Class 9 Solutions Maths helps students to build the basics and in-depth understanding of the fundamental concepts. In this chapter, students will learn about the Factorization of Algebraic Expressions using the factorization method. The process of factorization can be defined as the disintegration of a term into smaller factors, whereas algebraic expressions are built up of variables, integer constants and basic arithmetic operations of algebra.

The expert faculty at BYJU’S have formulated RD Sharma Solutions based on the marks allotted to each step of a problem. The main objective is to help students in solving problems effortlessly and make the subject easier to understand. Our subject experts have provided chapter-wise solutions in a precise manner to boost students’ academic performance. Students can download the RD Sharma Solutions to score high marks in the examination.

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

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Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 5 Factorization of Algebraic Expressions

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 Factorization of Algebraic Expressions

In the 5^th Chapter of Class 9 RD Sharma Solutions, students will study important topics, as listed below:

• Factorization by taking out the common factors
• Factorization by grouping the terms
• Factorization by making a perfect square
• Factorization by the difference of two squares
• Factorization of quadratic polynomials by splitting the middle term
• Factorization of algebraic expressions expressible as the sum or difference of two cubes

Students who want to gain proficiency in Mathematics are advised to practise textbook questions as many times as possible, using solutions provided at BYJU’S website. This also helps them to score high marks in the final examination.