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S.chand books class 8 maths solution chapter 14 Factorisation exercise 14 A

Exercise 14 A


Factorize completely by removing a monomial factor.

Question 1 

3y – 9

Sol :

= 3(y) + 3( -3)

The common factor is 3 , dividing each term by 3 we obtain other factor ( y – 3 )

= 3(y-3)

 


Question 12
 5x + 10

Sol :

= 5(x) + 5( 2)

The common factor is 5 , dividing each term by 3 we obtain other factor ( x + 2 )

= 5(x + 2)

 


Question 3

2x – 4

Sol :

= 2( x) + 2( -2 )

The common factor is 2 , dividing each term by 2 we obtain other factor ( x – 2 )

= 2( x – 2 )

 


Question 4

5m + 5n

Sol :

= 5( m ) + 5( n )

The common factor is 5 , dividing each term by 5 we obtain other factor ( m + n )

= 5( m + n )

 


Question 5

4a + 8b

Sol :

= 4( a ) + 4( 2b )

The common factor is 4 , dividing each term by 4 we obtain other factor ( a + 2b )

= 4( a + 2b )

 


Question 6

7x – 14y

Sol :

= 7( x ) + 7(-2y )

The common factor is 7 , dividing each term by 7 we obtain other factor ( x – 2y )

= 7( x – 2y )

 


Question 7

-3m – 15n

Sol :

= -3( m )+ {-3( 5n )]

The common factor is -3 , dividing each term by -3 we obtain other factor ( m + 5n )

= -3( m + 5n )

 


Question 8

-7p – 14q

Sol :

= -7( p ) + {-7( 2q )}

The common factor is -7 , dividing each term by -7 we obtain other factor ( p + 2q )

= -7( p + 2q )

 


Question 9

6x2 – 11x

Sol :

= x(6x) – x(11)

The common factor is x , dividing each term by x we obtain other factor (6x-11)

= x(6x-11)

 


Question 10

3y2 – 7y

Sol :

= y( 3y ) – y( 7 )

The common factor is y , dividing each term by y we obtain other factor ( 3y – 7 )

= y( 3y – 7 )

 


Question 11

ax + bx

Sol :

= x( a ) + x( b )

The common factor is x , dividing each term by x we obtain other factor ( a + b )

= x( a + b )

 


Question 12

x2y + xy2

Sol :

= x(xy) + (xy)y

The common factor is xy , dividing each term by xy we obtain other factor ( x + y )

= xy( x + y )

 


Question 13

4 + 12x2

Sol :

= 4( 1 ) + 4( 3x2 )

The common factor is 4 , dividing each term by 4 we obtain other factor ( 1 + 3x2 )

= 4( 1 + 3x2 )

 


Question 14

ax + ay + az

Sol :

= a( x ) + a( y ) + a( z )

The common factor is a , dividing each term by a we obtain other factor ( x + y + z )

= a( x + y + z )

 


Question 15

a3b + ab3

Sol :

= a2(ab) + (ab)b2

The common factor is ab , dividing each term by ab we obtain other factor (a2 + b2)

= ab(a2 + b2)

 


Question 16

x2y2 + x2

Sol :

= x2(y2)+ x2(1)

The common factor is x2 , dividing each term by x2 we obtain other factor (y2+1)

= x2(y2+1)

 


Question 17

p2 – 3pq +pq2

Sol :

= p(p) – p(3q) +p(q2)

The common factor is p , dividing each term by p we obtain other factor (p-3q+q2)

= p(p-3q+q2)

 


Question 18

7y3 – 5y2

Sol :

= (7y)y2 – (5)y2

The common factor is y , dividing each term by y we obtain other factor ( 7y – 5 )

= y2( 7y – 5 )

 


Question 19

6x3 -10x2

Sol :

= 2x2(3x) – 2x2(5)

The common factor is 2x2 , dividing each term by 2x2 we obtain other factor ( 3x – 5 )

= 2x2( 3x – 5 )

 


Question 20

2ab2 – 6bc + 6abc

Sol :

= 2b(ab) – 2b(3c) + 2b(4ac)

The common factor is 2b , dividing each term by 2b we obtain other factor ( ab – 3c + 4ac )

= 2b( ab – 3c + 4ac )

 


Question 21

12p5 + 16p4 – 20p3

Sol :

= 4p3(3p2)+ 4p3(4p) – 4p3(5)

The common factor is 4p3 , dividing each term by 4p3 we obtain other factor ( 3p2 + 4p – 5 )

= 4p3( 3p2 + 4p – 5 )


 

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