NCERT Solutions Class 8 Maths Chapter 14 Factorisation Exercise 14.2

 

Introduction :

In this exercise we will learn about Factorisation. Factorisation using identities. Factors of the form ( x + a ) ( x + b ). In general, for factorsing an algebraic expression of the type x² + px + q, we find two factors a and b of q ( i.e., the constant term ).

 

NCERT Class 9 Maths Chapter 5 Introduction to Euclid's Geometry :

• NCERT Class 8 Maths Understanding Factorisation Exercise 14.1
• NCERT Class 8 Maths Understanding Factorisation Exercise 14.2
• NCERT Class 8 Maths Understanding Factorisation Exercise 14.3
• NCERT Class 8 Maths Understanding Factorisation Exercise 14.4

NCERT Class 8 Maths Exercise 14.1 (Page-223)

Q1. Factories  the following  expressions.

( i ) a2 + 8a + 16 

Solution : 

We have a2 + 8a + 16 to factories 

Using the identity = ( x + y )2 = x2 + 2xy + y2

Here , x = a & b = 4 

= a2 + 2 × a × 4 + ( 4 ) 2

∴ So the required value are  ( a + 4 )2 . 

( ii ) p2 - 10p + 25

Solution :

We have  p2 - 10p + 25 to factories 

Using the identity = ( x - y )2 = x2- 2xy + y2  

Here , x = p & y = 5 

= p2 - 2 × p ×5 + ( 5 )2

∴ So the required value are ( p - 5 )2 .

( iii ) 25m2 + 30m + 9

Solution :

We have 25m2 + 30m + 9 to factories

Using the identity = ( x + y )2 = x2 + 2xy + y2

Here , x = 5  &  y = 3  

= ( 5m )2 + 2 × 5m × 3 + ( 3 )2

∴ So the required value are ( 5m + 3 )2 . 

( iv ) 49y2 + 84yz + 36z2

Solution :

We have 49y2 + 84yz + 36z2 to factories

Using the identity = ( x + y )2 = x2 + 2xy + y2

Here , x = 7y  &  y = 6z

= ( 7y )2 + 2 × 7y × 6z + ( 6z )2

∴ So the required value are ( 7y + 6z )2 . 

( v ) 4x2 - 8x + 4

Solution :

We have 4x2 - 8x + 4  to factories 

Using the identity = ( x - y )2 = x2- 2xy + y2  

Here , x = 2x & y = 2 

= ( 2 )2 - 2 × 2x ×2 + ( 2 )2

∴ So the required value are ( 2x - 2 )2 .

( vi ) 121b2 - 88bc + 16c2

Solution :

We have 121b2 - 88bc + 16c2  to factories 

Using the identity = ( x - y )2 = x2- 2xy + y2  

Here , x = 11b2  & y = 4c 2

= ( 11b )2 - 2 × 11b ×4c + ( 4c )2

∴ So the required value are ( 11b - 4c )2

( vii ) ( l+ m )2 - 4lm  ( Hint : Expand ( l + m)2 first)

Solution :

We have ( l + m )2 - 4lm  to factorise 

Expand ( l + m )2  by using identity = ( x + y )2 = x2 + 2xy + y2

Here x = l  & y = m

= l2 + 2lm + m2 - 4lm 

=l2 - 2lm + m2

∴ So the required value are ( l - m )2

( viii ) a4 + 2a2b2 + b4

Solution :

We have a4 + 2a2b2 + b4   to factories

Using the identity = ( x + y )2 = x2 + 2xy + y2

Here , x = a2  &  y = b2

= ( a2 )2 + 2 ×a2 × b2+ ( b2 )2

∴ So the required value are ( a2 +b2 )2 .

Q2. Factories .

( i ) 4p2 - 9q2

Solution : 

We have 4p2 - 9q2 to factories

Using the identity ( a + b )( a - b ) = a2 - b2

Here , a = 2p & b = 9q

= ( 2p )2 - ( 9q )2

∴ So the required value are ( 2p + 9q )( 2p - 9q ) . 

( ii ) 63a2 - 112b2

Solution : 

We have 63a2 - 112b2 to factories

Using the identity ( a + b )( a - b ) = a2 - b2

Here the common factor = 7 , a = 3a , b = 4b

= 7 ( 9a2 - 16b2 )

= 7 [ ( 3a )2 - ( 4b )2 ]

∴ So the required value are 7( 3a + 4b ) ( 3a - 4b ) .

( iii ) 49x2 - 36 

Solution : 

We have 49x2 - 36  to factories

Using the identity ( a + b )( a - b ) = a2 - b2

Here , a = 7x  & b = 6

= ( 7x )2 - ( 6 )2

∴ So the required value are ( 7x + 6 )( 7x - 6 ) . 

( iv ) 16x5 - 144x3

Solution :

We have 16x5 - 144x3  to factories

Using the identity ( a + b )( a - b ) = a2 - b2

Here the common factor = 16x3 , a = x2  , b = 9

= 16x3 ( x2 - 9 )

= 16x3 [ ( x )2 - ( 9 )2 ]

∴ So the required value are 16x3 ( x + 3 ) ( x - 3).

( v ) (l + m )2 - ( l - m )2

Solution :

We have (l + m )2 - ( l - m )2 to factorise

Using identity = ( x + y )2 = x2 + 2xy + y2 , ( x - y )2 = x2- 2xy + y2  

= [ ( l+m ) - ( l - m )] - [ ( l+m ) + ( l - m ) ]

= ( l + m - l + m ) ( l + m + l - m )

= ( 2m )( 2l )

= 4lm

( vi ) 9x2y2 - 16

Solution :

We have 9x2y2 - 16 to factories

Using the identity ( a + b )( a - b ) = a2 - b2

Here a = 3xy , b = 4

= ( 3xy )2 - ( 4 )2

∴ So the required value are ( 3xy + 4 ) ( 3xy - 4 ) .

( vii ) ( x2 - 2xy + y2 ) - z2

Solution :

We have ( x2 - 2xy + y2 ) - z2 to factories

Using the identity ( a + b )( a - b ) = a2 - b2

= ( x - y )2 - z2

∴ So the required value are ( x - y - z ) ( x - y + z )

( viii ) 25a2 - 4b2 + 25bc - 49c2

Solution : 

We have 25a2 - 4b2 + 25bc - 49c2  to factories

Using the identity ( a + b )( a - b ) = a2 - b2

= 25a2 - ( 4b2 - 25bc + 49c2)

= ( 5a )2 - ( 2b - 7c )2

∴ So the required value are  ( 5a - 2b + 7c ) ( 5a + 2b - 7c )

Q3. Factoies the expressions .

( i ) ax2 - bx

Solution :

We have ax2 - bx  to factories

Find the common factor

= ax ( x - b )

∴ So the required value are = ax ( x - b )

( ii ) 7p2 - 21q2

Solution :

We have 7p2 - 21q2  to factories

Find the common factor

= 7 ( p2 - 3q2 )

∴ So the required value are = 7 ( p2 - 3q2 ) 

( iii ) 2x3 + 2xy2 + 2xz2

Solution :

We have 7p2 - 21q2   to factories

Find the common factor

= 2x ( x2 + y2 + z2 )

∴ So the required value are = 2x ( x2 + y2 + z2 )

( iv ) am2 + bm2 + bn2 + an2

Solution :

We have am2 + bm2 + bn2 + an2  to factories

Find the common factor

= m ( a + b ) + n ( b + a )

=  ( m + n ) ( a + b )

∴ So the required value are = ( m + n ) ( a + b )

( v ) ( lm + l ) +  m + l

Solution : 

We have ( lm + l ) +  m + l  to factories

Find the common factor

= l ( m + l ) + 1 ( m + l )

= ( m + l ) ( l + 1 )

∴ So the required value are = ( m + l ) ( l + 1 )

( vi ) y ( y + z ) + 9 ( y + z )

Solution : 

We have y ( y + z ) + 9 ( y + z ) to factories

= ( y + z ) ( y + 9 )

∴ So the required value are = ( y + z ) ( y + 9 )

( vii ) 5y - 20y - 8z + 2yz

Solution : 

We have 5y2 - 20y - 8z + 2yz  to factories

Find the common factor

= 5y ( y - 4 ) + 2z ( -4 + z )

= ( y - 4 ) ( 5y + 2z )

∴ So the required value are = ( y - 4 ) ( 5y + 2z )

 

 

 

 

• NCERT Class 8 Maths Understanding Factorisation Exercise 14.1
• NCERT Class 8 Maths Understanding Factorisation Exercise 14.2
• NCERT Class 8 Maths Understanding Factorisation Exercise 14.3