{"id":545427,"date":"2021-10-04T09:58:55","date_gmt":"2021-10-04T09:58:55","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=545427"},"modified":"2021-10-05T09:08:22","modified_gmt":"2021-10-05T09:08:22","slug":"rd-sharma-solutions-for-class-9-maths-chapter-5-factorization-of-algebraic-expressions","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-9-maths-chapter-5-factorization-of-algebraic-expressions\/","title":{"rendered":"RD Sharma Solutions for Class 9 Maths Chapter 5\u2013Factorization of Algebraic Expressions"},"content":{"rendered":"\n

Class 9: Maths Chapter 5 solutions. Complete Class 9 Maths Chapter 5 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 5\u2013Factorization of Algebraic Expressions<\/h2>\n\n\n\n

RD Sharma 9th Maths Chapter 5, Class 9 Maths Chapter 5 solutions<\/p>\n\n\n\n

Exercise 5.1 Page No: 5.9<\/h3>\n\n\n\n

Question 1: Factorize x3<\/sup> + x \u2013 3x2<\/sup> \u2013 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

x3<\/sup> + x \u2013 3x2<\/sup> \u2013 3<\/p>\n\n\n\n

Here x is common factor in x3<\/sup> + x and \u2013 3 is common factor in \u2013 3x2<\/sup> \u2013 3<\/p>\n\n\n\n

x3<\/sup> \u2013 3x2<\/sup> + x \u2013 3<\/p>\n\n\n\n

x2<\/sup> (x \u2013 3) + 1(x \u2013 3)<\/p>\n\n\n\n

Taking ( x \u2013 3) common<\/p>\n\n\n\n

(x \u2013 3) (x2<\/sup> + 1)<\/p>\n\n\n\n

Therefore x3<\/sup> + x \u2013 3x2<\/sup> \u2013 3 = (x \u2013 3) (x2<\/sup> + 1)<\/p>\n\n\n\n

Question 2: Factorize a(a + b)3<\/sup> \u2013 3a2<\/sup>b(a + b)<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

a(a + b)3<\/sup> \u2013 3a2<\/sup>b(a + b)<\/p>\n\n\n\n

Taking a (a + b) as common factor<\/p>\n\n\n\n

= a(a + b) {(a + b)2<\/sup> \u2013 3ab}<\/p>\n\n\n\n

= a(a + b) {a2<\/sup> + b2<\/sup> + 2ab \u2013 3ab}<\/p>\n\n\n\n

= a(a + b) (a2<\/sup> + b2<\/sup> \u2013 ab)<\/p>\n\n\n\n

Question 3: Factorize x(x3<\/sup> \u2013 y3<\/sup>) + 3xy(x \u2013 y)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

x(x3<\/sup><\/strong> \u2013 y3<\/sup><\/strong>) + 3xy(x \u2013 y)<\/p>\n\n\n\n

= x(x \u2013 y) (x2<\/sup> + xy + y2<\/sup>) + 3xy(x \u2013 y)<\/p>\n\n\n\n

Taking x(x \u2013 y) as a common factor<\/p>\n\n\n\n

= x(x \u2013 y) (x2<\/sup> + xy + y2<\/sup> + 3y)<\/p>\n\n\n\n

= x(x \u2013 y) (x2<\/sup> + xy + y2<\/sup> + 3y)<\/p>\n\n\n\n

Question 4: Factorize a2<\/sup>x2<\/sup> + (ax2<\/sup> + 1)x + a<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

a2<\/sup>x2<\/sup> + (ax2<\/sup> + 1)x + a<\/p>\n\n\n\n

= a2<\/sup>x2<\/sup> + a + (ax2<\/sup> + 1)x<\/p>\n\n\n\n

= a(ax2<\/sup> + 1) + x(ax2<\/sup> + 1)<\/p>\n\n\n\n

= (ax2<\/sup> + 1) (a + x)<\/p>\n\n\n\n

Question 5: Factorize x2<\/sup> + y \u2013 xy \u2013 x<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

x2<\/sup> + y \u2013 xy \u2013 x<\/p>\n\n\n\n

= x2 <\/sup>\u2013 x \u2013 xy + y<\/p>\n\n\n\n

= x(x- 1) \u2013 y(x \u2013 1)<\/p>\n\n\n\n

= (x \u2013 1) (x \u2013 y)<\/p>\n\n\n\n

Question 6: Factorize x3<\/sup> \u2013 2x2<\/sup>y + 3xy2<\/sup> \u2013 6y3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

x3<\/sup> \u2013 2x2<\/sup>y + 3xy2<\/sup> \u2013 6y3<\/sup><\/p>\n\n\n\n

= x2<\/sup>(x \u2013 2y) + 3y2<\/sup>(x \u2013 2y)<\/p>\n\n\n\n

= (x \u2013 2y) (x2<\/sup> + 3y2<\/sup>)<\/p>\n\n\n\n

Question 7: Factorize 6ab \u2013 b2<\/sup> + 12ac \u2013 2bc<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

6ab \u2013 b2<\/sup> + 12ac \u2013 2bc<\/p>\n\n\n\n

= 6ab + 12ac \u2013 b2<\/sup> \u2013 2bc<\/p>\n\n\n\n

Taking 6a common from first two terms and \u2013b from last two terms<\/p>\n\n\n\n

= 6a(b + 2c) \u2013 b(b + 2c)<\/p>\n\n\n\n

Taking (b + 2c) common factor<\/p>\n\n\n\n

= (b + 2c) (6a \u2013 b)<\/p>\n\n\n\n

Question 8: Factorize (x2<\/sup> + 1\/x2<\/sup>) \u2013 4(x + 1\/x) + 6<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(x2<\/sup> + 1\/x2<\/sup>) \u2013 4(x + 1\/x) + 6<\/p>\n\n\n\n

= x2<\/sup> + 1\/x2<\/sup> \u2013 4x \u2013 4\/x + 4 + 2<\/p>\n\n\n\n

= x2<\/sup> + 1\/x2<\/sup> + 4 + 2 \u2013 4\/x \u2013 4x<\/p>\n\n\n\n

= (x2<\/sup>) + (1\/x) 2<\/sup> + ( -2 )2<\/sup> + 2x(1\/x) + 2(1\/x)(-2) + 2(-2)x<\/p>\n\n\n\n

As we know, x2<\/sup> + y2<\/sup> + z2<\/sup> + 2xy + 2yz + 2zx = (x+y+z) 2<\/sup><\/p>\n\n\n\n

So, we can write;<\/p>\n\n\n\n

= (x + 1\/x + (-2 )) 2<\/sup><\/p>\n\n\n\n

or (x + 1\/x \u2013 2) 2<\/sup><\/p>\n\n\n\n

Therefore, x2<\/sup> + 1\/x2<\/sup>) \u2013 4(x + 1\/x) + 6 = (x + 1\/x \u2013 2) 2<\/sup><\/p>\n\n\n\n

Question 9: Factorize x(x \u2013 2) (x \u2013 4) + 4x \u2013 8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

x(x \u2013 2) (x \u2013 4) + 4x \u2013 8<\/p>\n\n\n\n

= x(x \u2013 2) (x \u2013 4) + 4(x \u2013 2)<\/p>\n\n\n\n

= (x \u2013 2) [x(x \u2013 4) + 4]<\/p>\n\n\n\n

= (x \u2013 2) (x2<\/sup> \u2013 4x + 4)<\/p>\n\n\n\n

= (x \u2013 2) [x2<\/sup> \u2013 2 (x)(2) + (2) 2<\/sup>]<\/p>\n\n\n\n

= (x \u2013 2) (x \u2013 2) 2<\/sup><\/p>\n\n\n\n

= (x \u2013 2)3<\/sup><\/p>\n\n\n\n

Question 10: Factorize ( x + 2 ) ( x2<\/sup> + 25 ) \u2013 10x2<\/sup> \u2013 20x<\/strong><\/p>\n\n\n\n

Solution :<\/strong><\/p>\n\n\n\n

( x + 2) ( x2<\/sup> + 25) \u2013 10x ( x + 2 )<\/p>\n\n\n\n

Take ( x + 2 ) as common factor;<\/p>\n\n\n\n

= ( x + 2 )( x2 <\/sup><\/strong>+ 25 \u2013 10x)<\/p>\n\n\n\n

=( x + 2 ) ( x2 <\/sup><\/strong>\u2013 10x + 25)<\/p>\n\n\n\n

Expanding the middle term of ( x2<\/sup> \u2013 10x + 25 )<\/p>\n\n\n\n

=( x + 2 ) ( x2 <\/sup><\/strong>\u2013 5x \u2013 5x + 25 )<\/p>\n\n\n\n

=( x + 2 ){ x (x \u2013 5 ) \u2013 5 ( x \u2013 5 )}<\/p>\n\n\n\n

=( x + 2 )( x \u2013 5 )( x \u2013 5 )<\/p>\n\n\n\n

=( x + 2 )( x \u2013 5 )2<\/sup><\/p>\n\n\n\n

Therefore, ( x + 2) ( x2<\/sup> + 25) \u2013 10x ( x + 2 ) = ( x + 2 )( x \u2013 5 )2<\/sup><\/p>\n\n\n\n

Question 11: Factorize 2a2<\/sup> + 2\u221a6 ab + 3b2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

2a2<\/sup> + 2\u221a6 ab + 3b2<\/sup><\/p>\n\n\n\n

Above expression can be written as ( \u221a2a )2<\/sup> + 2 \u00d7 \u221a2a \u00d7 \u221a3b + ( \u221a3b)2<\/sup><\/p>\n\n\n\n

As we know, ( p + q ) 2 <\/sup>= p2 <\/sup>+ q2<\/sup> + 2pq<\/p>\n\n\n\n

Here p = \u221a2a and q = \u221a3b<\/p>\n\n\n\n

= (\u221a2a + \u221a3b )2<\/sup><\/p>\n\n\n\n

Therefore, 2a2<\/sup> + 2\u221a6 ab + 3b2<\/sup> = (\u221a2a + \u221a3b )2<\/sup><\/p>\n\n\n\n

Question 12: Factorize (a \u2013 b + c)2<\/sup> + (b \u2013 c + a) 2<\/sup> + 2(a \u2013 b + c) (b \u2013 c + a)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(a \u2013 b + c)2<\/sup> + ( b \u2013 c + a) 2<\/sup> + 2(a \u2013 b + c) (b \u2013 c + a)<\/p>\n\n\n\n

{Because p2 <\/sup>+ q2<\/sup> + 2pq = (p + q) 2<\/sup>}<\/p>\n\n\n\n

Here p = a \u2013 b + c and q = b \u2013 c + a<\/p>\n\n\n\n

= [a \u2013 b + c + b- c + a]2<\/sup><\/p>\n\n\n\n

= (2a)2<\/sup><\/p>\n\n\n\n

= 4a2<\/sup><\/p>\n\n\n\n

Question 13: Factorize a2<\/sup> + b2 <\/sup>+ 2( ab+bc+ca )<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

a2<\/sup> + b2 <\/sup>+ 2ab + 2bc + 2ca<\/p>\n\n\n\n

As we know, p2 <\/sup>+ q2<\/sup> + 2pq = (p + q) 2<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= ( a+b)2<\/sup> + 2bc + 2ca<\/p>\n\n\n\n

= ( a+b)2<\/sup> + 2c( b + a )<\/p>\n\n\n\n

Or ( a+b)2<\/sup> + 2c( a + b )<\/p>\n\n\n\n

Take ( a + b ) as common factor;<\/p>\n\n\n\n

= ( a + b )( a + b + 2c )<\/p>\n\n\n\n

Therefore, a2<\/sup> + b2 <\/sup>+ 2ab + 2bc + 2ca = ( a + b )( a + b + 2c )<\/p>\n\n\n\n

Question 14: Factorize 4(x-y) 2<\/sup> \u2013 12(x \u2013 y)(x + y) + 9(x + y)2<\/sup><\/strong><\/p>\n\n\n\n

Solution :<\/strong><\/p>\n\n\n\n

Consider ( x \u2013 y ) = p, ( x + y ) = q<\/p>\n\n\n\n

= 4p2<\/sup> \u2013 12pq + 9q2<\/sup><\/p>\n\n\n\n

Expanding the middle term, -12 = -6 -6 also 4\u00d7 9=-6 \u00d7 -6<\/p>\n\n\n\n

= 4p2<\/sup> \u2013 6pq \u2013 6pq + 9q2<\/sup><\/p>\n\n\n\n

=2p( 2p \u2013 3q ) -3q( 2p \u2013 3q )<\/p>\n\n\n\n

= ( 2p \u2013 3q ) ( 2p \u2013 3q )<\/p>\n\n\n\n

= ( 2p \u2013 3q )2<\/sup><\/p>\n\n\n\n

Substituting back p = x \u2013 y and q = x + y;<\/p>\n\n\n\n

= [2( x-y ) \u2013 3( x+y)]2 <\/sup>= [ 2x \u2013 2y \u2013 3x \u2013 3y ] 2<\/sup><\/p>\n\n\n\n

= (2x-3x-2y-3y ) 2<\/sup><\/p>\n\n\n\n

=[ -x \u2013 5y] 2<\/sup><\/p>\n\n\n\n

=[( -1 )( x+5y )] 2<\/sup><\/p>\n\n\n\n

=( x+5y ) 2<\/sup><\/p>\n\n\n\n

Therefore, 4(x-y) 2<\/sup> \u2013 12(x \u2013 y)(x + y) + 9(x + y)2<\/sup> = ( x+5y )2<\/sup><\/p>\n\n\n\n

Question 15: Factorize a2<\/sup> \u2013 b2 <\/sup>+ 2bc \u2013 c2<\/sup><\/strong><\/p>\n\n\n\n

Solution :<\/strong><\/p>\n\n\n\n

a2<\/sup> \u2013 b2 <\/sup>+ 2bc \u2013 c2<\/sup><\/p>\n\n\n\n

As we know, ( a-b)2<\/sup> = a2 <\/sup>+ b2 <\/sup>\u2013 2ab<\/p>\n\n\n\n

= a2 <\/sup>\u2013 ( b \u2013 c) 2<\/sup><\/p>\n\n\n\n

Also we know, a2 <\/sup>\u2013 b2 <\/sup>= ( a+b)( a-b)<\/p>\n\n\n\n

= ( a + b \u2013 c )( a \u2013 ( b \u2013 c ))<\/p>\n\n\n\n

= ( a + b \u2013 c )( a \u2013 b + c )<\/p>\n\n\n\n

Therefore, a2<\/sup> \u2013 b2 <\/sup>+ 2bc \u2013 c2 <\/sup>=( a + b \u2013 c )( a \u2013 b + c )<\/p>\n\n\n\n

Question 16: Factorize a2<\/sup> + 2ab + b2<\/sup> \u2013 c2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

a2<\/sup> + 2ab + b2<\/sup> \u2013 c2<\/sup><\/p>\n\n\n\n

= (a2<\/sup> + 2ab + b2<\/sup>) \u2013 c2<\/sup><\/p>\n\n\n\n

= (a + b)2<\/sup> \u2013 (c) 2<\/sup><\/p>\n\n\n\n

We know, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

= (a + b + c) (a + b \u2013 c)<\/p>\n\n\n\n

Therefore a2<\/sup> + 2ab + b2<\/sup> \u2013 c2 <\/sup>= (a + b + c) (a + b \u2013 c)<\/p>\n\n\n\n

\n\n\n\n

Exercise 5.2 Page No: 5.13<\/h4>\n\n\n\n

Factorize each of the following expressions:<\/strong><\/p>\n\n\n\n

Question 1: p3<\/sup> + 27<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

p3<\/sup> + 27<\/p>\n\n\n\n

= p3<\/sup> + 33<\/sup>[using a3<\/sup> + b3 <\/sup>= (a + b)(a2<\/sup> \u2013ab + b2<\/sup>)]<\/p>\n\n\n\n

= (p + 3)(p\u00b2 \u2013 3p \u2013 9)<\/p>\n\n\n\n

Therefore, p3<\/sup> + 27 = (p + 3)(p\u00b2 \u2013 3p \u2013 9)<\/p>\n\n\n\n

Question 2: y3<\/sup> + 125<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

y3<\/sup> + 125<\/p>\n\n\n\n

= y3<\/sup> + 53<\/sup><\/strong>[using a3<\/sup> + b3 <\/sup>= (a + b)(a2<\/sup> \u2013ab + b2<\/sup>)]<\/p>\n\n\n\n

= (y+5)(y2<\/sup> \u2212 5y + 52<\/sup>)<\/p>\n\n\n\n

= (y + 5)(y2 <\/sup>\u2212 5y + 25)<\/p>\n\n\n\n

Therefore, y3<\/sup> + 125 = (y + 5)(y2 <\/sup>\u2212 5y + 25)<\/p>\n\n\n\n

Question 3: 1 \u2013 27a3<\/sup><\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

= (1)3<\/sup> \u2212(3a) 3<\/sup>[using a3<\/sup> \u2013 b3 <\/sup>= (a \u2013 b)(a2<\/sup> + ab + b2<\/sup>)]<\/p>\n\n\n\n

= (1\u2212 3a)(12<\/sup> + 1\u00d73a + (3a) 2<\/sup>)<\/p>\n\n\n\n

= (1\u22123a)(1 + 3a + 9a2<\/sup>)<\/p>\n\n\n\n

Therefore, 1\u221227a3<\/sup> = (1\u22123a)(1 + 3a+ 9a2<\/sup>)<\/p>\n\n\n\n

Question 4: 8x3<\/sup>y3<\/sup> + 27a3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

8x3<\/sup>y3<\/sup> + 27a3<\/sup><\/p>\n\n\n\n

= (2xy) 3<\/sup> + (3a) 3<\/sup>[using a3<\/sup> + b3 <\/sup>= (a + b)(a2<\/sup> \u2013ab + b2<\/sup>)]<\/p>\n\n\n\n

= (2xy +3a)((2xy)2<\/sup>\u22122xy\u00d73a+(3a) 2<\/sup>)<\/p>\n\n\n\n

= (2xy+3a)(4x2<\/sup>y2 <\/sup>\u22126xya + 9a2<\/sup>)<\/p>\n\n\n\n

Question 5: 64a3<\/sup> \u2212 b3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

64a3<\/sup> \u2212 b3<\/sup><\/p>\n\n\n\n

= (4a)3<\/sup>\u2212b3<\/sup>[using a3<\/sup> \u2013 b3 <\/sup>= (a \u2013 b)(a2<\/sup> + ab + b2<\/sup>)]<\/p>\n\n\n\n

= (4a\u2212b)((4a)2<\/sup> + 4a\u00d7b + b2<\/sup>)<\/p>\n\n\n\n

=(4a\u2212b)(16a2 <\/sup>+4ab+b2<\/sup>)<\/p>\n\n\n\n

Question 6: x3<\/sup> \/ 216 \u2013 8y3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

x3<\/sup> \/ 216 \u2013 8y3<\/sup><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 7: 10x4 <\/sup>y \u2013 10xy4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

10x4 <\/sup>y \u2013 10xy4<\/sup><\/p>\n\n\n\n

= 10xy(x3 <\/sup>\u2212 y3<\/sup>)[using a3<\/sup> \u2013 b3 <\/sup>= (a \u2013 b)(a2<\/sup> + ab + b2<\/sup>)]<\/p>\n\n\n\n

= 10xy (x\u2212y)(x2 <\/sup>+ xy + y2<\/sup>)<\/p>\n\n\n\n

Therefore, 10x4 <\/sup>y \u2013 10xy4 <\/sup>= 10xy (x\u2212y)(x2 <\/sup>+ xy + y2<\/sup>)<\/p>\n\n\n\n

Question 8: 54x6 <\/sup>y + 2x3<\/sup>y4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

54x6 <\/sup>y + 2x3<\/sup>y4<\/sup><\/p>\n\n\n\n

= 2x3<\/sup>y(27x3<\/sup> +y3<\/sup>)<\/p>\n\n\n\n

= 2x3<\/sup>y((3x) 3 <\/sup>+ y3<\/sup>)[using a3<\/sup> + b3 <\/sup>= (a + b)(a2<\/sup> \u2013 ab + b2<\/sup>)]<\/p>\n\n\n\n

= 2x3<\/sup>y {(3x+y) ((3x)2<\/sup>\u22123xy+y2<\/sup>)}<\/p>\n\n\n\n

=2x3<\/sup>y(3x+y)(9x2<\/sup> \u2212 3xy + y2<\/sup>)<\/p>\n\n\n\n

Question 9: 32a3 <\/sup>+ 108b3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

32a3 <\/sup>+ 108b3<\/sup><\/p>\n\n\n\n

= 4(8a3<\/sup> + 27b3<\/sup>)<\/p>\n\n\n\n

= 4((2a) 3<\/sup>+(3b) 3<\/sup>)[using a3<\/sup> + b3 <\/sup>= (a + b)(a2<\/sup> \u2013 ab + b2<\/sup>)]<\/p>\n\n\n\n

= 4[(2a+3b)((2a)2<\/sup>\u22122a\u00d73b+(3b) 2<\/sup>)]<\/p>\n\n\n\n

= 4(2a+3b)(4a2 <\/sup>\u2212 6ab + 9b2<\/sup>)<\/p>\n\n\n\n

Question 10: (a\u22122b)3<\/sup> \u2212 512b3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(a\u22122b)3<\/sup> \u2212 512b3<\/sup><\/p>\n\n\n\n

= (a\u22122b)3 <\/sup>\u2212(8b) 3<\/sup>[using a3<\/sup> \u2013 b3 <\/sup>= (a \u2013 b)(a2<\/sup> + ab + b2<\/sup>)]<\/p>\n\n\n\n

= (a \u22122b\u22128b) {(a\u22122b)2 <\/sup>+ (a\u22122b)8b + (8b) 2<\/sup>}<\/p>\n\n\n\n

=(a \u221210b)(a2 <\/sup>+ 4b2 <\/sup>\u2212 4ab + 8ab \u2212 16b2 <\/sup>+ 64b2<\/sup>)<\/p>\n\n\n\n

=(a\u221210b)(a2 <\/sup>+ 52b2 <\/sup>+ 4ab)<\/p>\n\n\n\n

Question 11: (a+b)3<\/sup> \u2212 8(a\u2212b)3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(a+b)3<\/sup> \u2212 8(a\u2212b)3<\/sup><\/p>\n\n\n\n

= (a+b)3 <\/sup>\u2212 [2(a\u2212b)]3<\/sup><\/p>\n\n\n\n

= (a+b)3<\/sup> \u2212 [2a\u22122b] 3<\/sup>[using p3<\/sup> \u2013 q3 <\/sup>= (p \u2013 q)(p2<\/sup> + pq + q2<\/sup>)]<\/p>\n\n\n\n

Here p = a+b and q = 2a\u22122b<\/p>\n\n\n\n

= (a+b\u2212(2a\u22122b))((a+b)2<\/sup>+(a+b)(2a\u22122b)+(2a\u22122b) 2<\/sup>)<\/p>\n\n\n\n

=(a+b\u22122a+2b)(a2<\/sup>+b2<\/sup>+2ab+(a+b)(2a\u22122b)+(2a\u22122b) 2<\/sup>)<\/p>\n\n\n\n

=(a+b\u22122a+2b)(a2<\/sup>+b2<\/sup>+2ab+2a2<\/sup>\u22122ab+2ab\u22122b2<\/sup>+(2a\u22122b) 2<\/sup>)<\/p>\n\n\n\n

=(3b\u2212a)(3a2<\/sup>+2ab\u2212b2<\/sup>+(2a\u22122b) 2<\/sup>)<\/p>\n\n\n\n

=(3b\u2212a)(3a2<\/sup>+2ab\u2212b2<\/sup>+4a2<\/sup>+4b2<\/sup>\u22128ab)<\/p>\n\n\n\n

=(3b\u2212a)(3a2<\/sup>+4a2<\/sup>\u2212b2<\/sup>+4b2<\/sup>\u22128ab+2ab)<\/p>\n\n\n\n

=(3b\u2212a)(7a2<\/sup>+3b2<\/sup>\u22126ab)<\/p>\n\n\n\n

Question 12: (x+2)3 <\/sup>+ (x\u22122) 3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(x+2)3 <\/sup>+ (x\u22122) 3<\/sup>[using p3<\/sup> + q3 <\/sup>= (p + q)(p2<\/sup> \u2013 pq + q2<\/sup>)]<\/p>\n\n\n\n

Here p = x + 2 and q = x \u2013 2<\/p>\n\n\n\n

= (x+2+x\u22122)((x+2)2<\/sup>\u2212(x+2)(x\u22122)+(x\u22122) 2<\/sup>)<\/p>\n\n\n\n

=2x(x2 <\/sup>+4x+4\u2212(x+2)(x\u22122)+x2<\/sup>\u22124x+4)[ Using : (a+b)(a\u2212b) = a2<\/sup>\u2212b2<\/sup> ]<\/p>\n\n\n\n

= 2x(2x2 <\/sup>+ 8 \u2212 (x2 <\/sup>\u2212 22<\/sup>))<\/p>\n\n\n\n

= 2x(2x2 <\/sup>+8 \u2212 x2 <\/sup>+ 4)<\/p>\n\n\n\n

= 2x(x2 <\/sup>+ 12)<\/p>\n\n\n\n

\n\n\n\n

Exercise 5.3 Page No: 5.17<\/h4>\n\n\n\n

Question 1: Factorize 64a3<\/sup> + 125b3<\/sup> + 240a2<\/sup>b + 300ab2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

64a3<\/sup> + 125b3<\/sup> + 240a2<\/sup>b + 300ab2<\/sup><\/p>\n\n\n\n

= (4a)3<\/sup> + (5b) 3 <\/sup>+ 3(4a)2<\/sup>(5b) + 3(4a)(5b)2<\/sup> , which is similar to a3<\/sup> + b3 <\/sup>+ 3a2<\/sup>b + 3ab2<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> + b3 <\/sup>+ 3a2<\/sup>b + 3ab2 <\/sup>= (a+b)3<\/sup>]<\/p>\n\n\n\n

= (4a+5b)3<\/sup><\/p>\n\n\n\n

Question 2: Factorize 125x3<\/sup> \u2013 27y3<\/sup> \u2013 225x2<\/sup>y + 135xy2<\/sup><\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

125x3<\/sup> \u2013 27y3<\/sup> \u2013 225x2<\/sup>y + 135xy2<\/sup><\/p>\n\n\n\n

Above expression can be written as (5x)3<\/sup>\u2212(3y) 3<\/sup>\u22123(5x)2<\/sup>(3y) + 3(5x)(3y)2<\/sup><\/p>\n\n\n\n

Using: a3<\/sup> \u2212 b3<\/sup> \u2212 3a2<\/sup>b + 3ab2<\/sup> = (a\u2212b)3<\/sup><\/p>\n\n\n\n

= (5x \u2212 3y)3<\/sup><\/p>\n\n\n\n

Question 3: Factorize 8\/27 x3<\/sup> + 1 + 4\/3 x2<\/sup> + 2x<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

8\/27 x3<\/sup> + 1 + 4\/3 x2<\/sup> + 2x<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 4: Factorize 8x3<\/sup> + 27y3<\/sup> + 36x2<\/sup>y + 54xy2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

8x3<\/sup> + 27y3<\/sup> + 36x2<\/sup>y + 54xy2<\/sup><\/p>\n\n\n\n

Above expression can be written as (2x)3<\/sup> + (3y) 3 <\/sup>+ 3\u00d7(2x)2<\/sup>\u00d73y + 3\u00d7(2x)(3y)2<\/sup><\/p>\n\n\n\n

Which is similar to a\u00b3 + b\u00b3 + 3a\u00b2b + 3ab\u00b2 = (a + b) \u00b3]<\/p>\n\n\n\n

Here a = 2x and b = 3y<\/p>\n\n\n\n

= (2x+3y)3<\/sup><\/p>\n\n\n\n

Therefore, 8x3<\/sup> + 27y3<\/sup> + 36x2<\/sup>y + 54xy2<\/sup> = (2x+3y)3<\/sup><\/p>\n\n\n\n

Question 5: Factorize a3<\/sup> \u2212 3a2<\/sup>b + 3ab2<\/sup> \u2212 b3<\/sup> + 8<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

a3<\/sup> \u2212 3a2<\/sup>b + 3ab2<\/sup> \u2212 b3<\/sup> + 8<\/p>\n\n\n\n

Using: a3<\/sup> \u2212 b3<\/sup> \u2212 3a2<\/sup>b + 3ab2<\/sup> = (a\u2212b)3<\/sup><\/p>\n\n\n\n

= (a\u2212b)3<\/sup> + 23<\/sup><\/p>\n\n\n\n

Again , Using: a3<\/sup> + b3<\/sup> =(a + b)(a2<\/sup> \u2013 ab + b2<\/sup>)]<\/p>\n\n\n\n

=(a\u2212b+2)((a\u2212b)2<\/sup>\u2212(a\u2212b) \u00d7 2 + 22<\/sup>)<\/p>\n\n\n\n

=(a\u2212b+2)(a2<\/sup>+b2<\/sup>\u22122ab\u22122(a\u2212b)+4)<\/p>\n\n\n\n

=(a\u2212b+2)(a2<\/sup>+b2<\/sup>\u22122ab\u22122a+2b+4)<\/p>\n\n\n\n

a3<\/sup> \u2212 3a2<\/sup>b + 3ab2<\/sup> \u2212 b3<\/sup> + 8 =(a\u2212b+2)(a2<\/sup>+b2<\/sup>\u22122ab\u22122a+2b+4)<\/p>\n\n\n\n

\n\n\n\n

Exercise 5.4 Page No: 5.22<\/h4>\n\n\n\n

Factorize each of the following expressions:<\/strong><\/p>\n\n\n\n

Question 1: a3<\/sup> + 8b3 <\/sup>+ 64c3 <\/sup>\u2212 24abc<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

a3<\/sup> + 8b3 <\/sup>+ 64c3 <\/sup>\u2212 24abc<\/p>\n\n\n\n

= (a)3<\/sup> + (2b) 3 <\/sup>+ (4c) 3<\/sup>\u2212 3\u00d7a\u00d72b\u00d74c[Using a3<\/sup>+b3<\/sup>+c3<\/sup>\u22123abc = (a+b+c)(a2<\/sup>+b2<\/sup>+c2<\/sup>\u2212ab\u2212bc\u2212ca)]<\/p>\n\n\n\n

= (a+2b+4c)(a2<\/sup>+(2b)2<\/sup> + (4c)2<\/sup>\u2212a\u00d72b\u22122b\u00d74c\u22124c\u00d7a)<\/p>\n\n\n\n

= (a+2b+4c)(a2 <\/sup>+4b2 <\/sup>+16c2 <\/sup>\u22122ab\u22128bc\u22124ac)<\/p>\n\n\n\n

Therefore, a3<\/sup> + 8b3 <\/sup>+ 64c3 <\/sup>\u2212 24abc = (a+2b+4c)(a2 <\/sup>+4b2 <\/sup>+16c2 <\/sup>\u22122ab\u22128bc\u22124ac)<\/p>\n\n\n\n

Question 2: x 3 <\/sup>\u2212 8y 3<\/sup>+ 27z3<\/sup> + 18xyz<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

= x3<\/sup> \u2212 (2y) 3<\/sup> + (3z) 3<\/sup> \u2212 3\u00d7x\u00d7(\u22122y)(3z)<\/p>\n\n\n\n

= (x + (\u22122y) + 3z) (x2<\/sup> + (\u22122y)2 <\/sup>+ (3z) 2 <\/sup>\u2212x(\u22122y)\u2212(\u22122y)(3z)\u22123z(x))[using a3<\/sup>+b3<\/sup>+c3<\/sup>\u22123abc = (a+b+c)(a2<\/sup>+b2<\/sup>+c2<\/sup>\u2212ab\u2212bc\u2212ca)]<\/p>\n\n\n\n

=(x \u22122y + 3z)(x2<\/sup> + 4y2 <\/sup>+ 9 z2 <\/sup>+ 2xy + 6yz \u2212 3zx)<\/p>\n\n\n\n

Question 3: 27x 3 <\/sup>\u2212 y 3<\/sup>\u2013 z3<\/sup> \u2013 9xyz<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

27x 3 <\/sup>\u2212 y 3<\/sup>\u2013 z3<\/sup> \u2013 9xyz<\/p>\n\n\n\n

= (3x) 3 <\/sup>\u2212 y 3<\/sup>\u2013 z3<\/sup> \u2013 3(3xyz)[Using a3<\/sup> + b3<\/sup> + c3 <\/sup>\u22123abc = (a + b + c)(a2<\/sup>+b2<\/sup>+c2<\/sup>\u2212ab\u2212bc\u2212ca)]<\/p>\n\n\n\n

Here a = 3x, b = -y and c = -z<\/p>\n\n\n\n

= (3x \u2013 y \u2013 z){ (3x)2<\/sup> + (- y)2<\/sup> + (\u2013 z)2<\/sup> + 3xy \u2013 yz + 3xz)}<\/p>\n\n\n\n

= (3x \u2013 y \u2013 z){ 9x2<\/sup> + y2<\/sup> + z2<\/sup> + 3xy \u2013 yz + 3xz)}<\/p>\n\n\n\n

Question 4: 1\/27 x3<\/sup> \u2212 y3<\/sup> + 125z3<\/sup> + 5xyz<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

1\/27 x3<\/sup> \u2212 y3<\/sup> + 125z3<\/sup> + 5xyz<\/p>\n\n\n\n

= (x\/3)3<\/sup>+(\u2212y)3<\/sup> +(5z)3<\/sup> \u2013 3 x\/3 (\u2212y)(5z)[Using a3<\/sup> + b3<\/sup> + c3 <\/sup>\u22123abc = (a + b + c)(a2<\/sup>+b2<\/sup>+c2<\/sup>\u2212ab\u2212bc\u2212ca)]<\/p>\n\n\n\n

= (x\/3 + (\u2212y) + 5z)((x\/3)2<\/sup> + (\u2212y)2<\/sup> + (5z) 2 <\/sup>\u2013x\/3(\u2212y) \u2212 (\u2212y)5z\u22125z(x\/3))<\/p>\n\n\n\n

= (x\/3 \u2212y + 5z) (x^2\/9 + y2<\/sup> + 25z2 <\/sup>+ xy\/3 + 5yz \u2013 5zx\/3)<\/p>\n\n\n\n

Question 5: 8x3<\/sup> + 27y3 <\/sup>\u2212 216z3 <\/sup>+ 108xyz<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

8x3<\/sup> + 27y3 <\/sup>\u2212 216z3 <\/sup>+ 108xyz<\/p>\n\n\n\n

= (2x) 3 <\/sup>+ (3y) 3 <\/sup>+(\u22126y) 3 <\/sup>\u22123(2x)(3y)(\u22126z)<\/p>\n\n\n\n

= (2x+3y+(\u22126z)){ (2x)2<\/sup>+(3y) 2<\/sup>+(\u22126z) 2 <\/sup>\u22122x\u00d73y\u22123y(\u22126z)\u2212(\u22126z)2x}<\/p>\n\n\n\n

= (2x+3y\u22126z) {4x2 <\/sup>+9y2 <\/sup>+36z2 <\/sup>\u22126xy + 18yz + 12zx}<\/p>\n\n\n\n

Question 6: 125 + 8x3<\/sup> \u2212 27y3<\/sup> + 90xy<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

125 + 8x3<\/sup> \u2212 27y3<\/sup> + 90xy<\/p>\n\n\n\n

= (5)3<\/sup> + (2x) 3 <\/sup>+(\u22123y) 3 <\/sup>\u22123\u00d75\u00d72x\u00d7(\u22123y)<\/p>\n\n\n\n

= (5+2x+(\u22123y)) (52<\/sup> +(2x) 2 <\/sup>+(\u22123y) 2 <\/sup>\u22125(2x)\u22122x(\u22123y)\u2212(\u22123y)5)<\/p>\n\n\n\n

= (5+2x\u22123y)(25+4x2 <\/sup>+9y2 <\/sup>\u221210x+6xy+15y)<\/p>\n\n\n\n

Question 7: (3x\u22122y)3<\/sup> + (2y\u22124z) 3<\/sup> + (4z\u22123x) 3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(3x\u22122y)3<\/sup> + (2y\u22124z) 3<\/sup> + (4z\u22123x) 3<\/sup><\/p>\n\n\n\n

Let (3x\u22122y) = a, (2y\u22124z) = b , (4z\u22123x) = c<\/p>\n\n\n\n

a + b + c= 3x\u22122y+2y\u22124z+4z\u22123x = 0<\/p>\n\n\n\n

We know, a3<\/sup> + b3<\/sup> + c3 <\/sup>\u22123abc = (a + b + c)(a2<\/sup>+b2<\/sup>+c2<\/sup>\u2212ab\u2212bc\u2212ca)<\/p>\n\n\n\n

\u21d2 a3<\/sup> + b3<\/sup> + c3 <\/sup>\u22123abc = 0<\/p>\n\n\n\n

or a3<\/sup> + b3<\/sup> + c3 <\/sup>=3abc<\/p>\n\n\n\n

\u21d2 (3x\u22122y)3<\/sup> + (2y\u22124z) 3<\/sup> + (4z\u22123x) 3<\/sup> = 3(3x\u22122y)(2y\u22124z)(4z\u22123x)<\/p>\n\n\n\n

Question 8: (2x\u22123y)3<\/sup> + (4z\u22122x) 3<\/sup> + (3y\u22124z) 3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(2x\u22123y)3<\/sup> + (4z\u22122x) 3<\/sup> + (3y\u22124z) 3<\/sup><\/p>\n\n\n\n

Let 2x \u2013 3y = a , 4z \u2013 2x = b , 3y \u2013 4z = c<\/p>\n\n\n\n

a + b + c= 2x \u2013 3y + 4z \u2013 2x + 3y \u2013 4z = 0<\/p>\n\n\n\n

We know, a3<\/sup> + b3<\/sup> + c3 <\/sup>\u22123abc = (a + b + c)(a2<\/sup>+b2<\/sup>+c2<\/sup>\u2212ab\u2212bc\u2212ca)<\/p>\n\n\n\n

\u21d2 a3<\/sup> + b3<\/sup> + c3 <\/sup>\u22123abc = 0<\/p>\n\n\n\n

(2x\u22123y)3<\/sup> + (4z\u22122x) 3<\/sup> + (3y\u22124z) 3 <\/sup>= 3(2x\u22123y)(4z\u22122x)(3y\u22124z)<\/p>\n\n\n\n

\n\n\n\n

Exercise VSAQs Page No: 5.24<\/h4>\n\n\n\n

Question 1: Factorize x4<\/sup> + x2<\/sup> + 25<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

x4<\/sup> + x2<\/sup> + 25<\/p>\n\n\n\n

= (x2<\/sup>) 2 <\/sup>+ 52 <\/sup>+ x2<\/sup>[using a2<\/sup> + b2<\/sup> = (a + b) 2<\/sup> \u2013 2ab ]<\/p>\n\n\n\n

= (x2 <\/sup>+5) 2 <\/sup>\u22122(x2 <\/sup>) (5) + x2<\/sup><\/p>\n\n\n\n

=(x2 <\/sup>+5) 2 <\/sup>\u221210x2 <\/sup>+ x2<\/sup><\/p>\n\n\n\n

=(x2 <\/sup>+ 5) 2 <\/sup>\u2212 9x2<\/sup><\/p>\n\n\n\n

=(x2 <\/sup>+ 5) 2 <\/sup>\u2212 (3x) 2<\/sup>[using a2<\/sup> \u2013 b2<\/sup> = (a + b)(a \u2013 b ]<\/p>\n\n\n\n

= (x 2 <\/sup>+ 3x + 5)(x2 <\/sup>\u2212 3x + 5)<\/p>\n\n\n\n

Question 2: Factorize x2<\/sup> \u2013 1 \u2013 2a \u2013 a2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

x2<\/sup> \u2013 1 \u2013 2a \u2013 a2<\/sup><\/p>\n\n\n\n

x2<\/sup> \u2013 (1 + 2a + a2<\/sup> )<\/p>\n\n\n\n

x2<\/sup> \u2013 (a + 1)2<\/sup><\/p>\n\n\n\n

(x \u2013 (a + 1)(x + (a + 1)<\/p>\n\n\n\n

(x \u2013 a \u2013 1)(x + a + 1)[using a2<\/sup> \u2013 b2<\/sup> = (a + b)(a \u2013 b) and (a + b)^2 = a^2 + b^2 + 2ab ]<\/p>\n\n\n\n

Question 3: If a + b + c =0, then write the value of a3<\/sup> + b3<\/sup> + c3<\/sup>.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know, a3<\/sup> + b3<\/sup> + c3<\/sup> \u2013 3abc = (a + b +c ) (a2<\/sup> + b2 <\/sup>+ c2 <\/sup>\u2013 ab \u2013 bc \u2212 ca)<\/p>\n\n\n\n

Put a + b + c =0<\/p>\n\n\n\n

This implies<\/p>\n\n\n\n

a3<\/sup> + b3<\/sup> + c3<\/sup> = 3abc<\/p>\n\n\n\n

Question 4: If a2<\/sup> + b2<\/sup> + c2<\/sup> = 20 and a + b + c =0, find ab + bc + ca.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know, (a+b+c)\u00b2 = a\u00b2 + b\u00b2 + c\u00b2 + 2(ab + bc + ca)<\/p>\n\n\n\n

0 = 20 + 2(ab + bc + ca)<\/p>\n\n\n\n

-10 = ab + bc + ca<\/p>\n\n\n\n

Or ab + bc + ca = -10<\/p>\n\n\n\n

Question 5: If a + b + c = 9 and ab + bc + ca = 40, find a2<\/sup> + b2<\/sup> + c2<\/sup> .<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know, (a+b+c)\u00b2 = a\u00b2 + b\u00b2 + c\u00b2 + 2(ab + bc + ca)<\/p>\n\n\n\n

92<\/sup> = a\u00b2 + b\u00b2 + c\u00b2 + 2(40)<\/p>\n\n\n\n

81 = a\u00b2 + b\u00b2 + c\u00b2 + 80<\/p>\n\n\n\n

\u21d2 a\u00b2 + b\u00b2 + c\u00b2 = 1<\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 5: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 5\u2013Factorization of Algebraic Expressions<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 9 Maths Chapter 5\u2013Factorization of Algebraic Expressions PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n