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

Exercise 14 A

Factorize completely by removing a monomial factor.

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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)

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Question 2
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)

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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 )

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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 )

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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 )

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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 )

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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 )

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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 )

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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)

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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 )

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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 )

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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 )

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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)

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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 )

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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)

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Question 16

x²y² + x²

Sol :

= x²(y²)+ x²(1)

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

= x²(y²+1)

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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)

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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 )

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Question 19

6x³ -10x²

Sol :

= 2x²(3x) - 2x²(5)

The common factor is 2x² , dividing each term by 2x² weobtain other factor ( 3x - 5 )

= 2x²( 3x - 5 )

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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 )

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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 )

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

Exercise 14 A

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