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Q.

Factorise the following expressions. 

(i) 7x – 42                                         (ii) 6p – 12q 

(iii) 7a2 + 14a                                   (iv) – 16 z + 20 z

(v) 20 l2 m + 30 a l m                      (vi) 5 x2 y – 15 xy

(vii) 10 a 2 – 15 b2 + 20 c2              (viii) – 4 a2 + 4 ab – 4 ca 

 (ix) x 2 y z + x y2 z + x y z            (x) a x2 y + b x y2 + c x y z

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Detailed Solution

We know that, Factorization means finding out the factors of the given algebraic expression

(i) 7x - 42 : 

let 7x, 42 can be written as 

7x = 7 × x 

42 = 2 × 3 × 7 

The common factors is given as 7

Hence, we get Algebraic expression as 

7x - 42 = (7 × x)  -(2 × 3 × 7 )

= 7 (x - 6)

 (ii) 6p – 12q :

let 6p, 12q can be written as 

6p = 2 × 3 × p

12q = 2 × 2 × 3 × q

The common factors are given as 2,3

Hence, we get  Algebraic expression as 

6p – 12q = ( 2 × 3 × p) -( 2 × 2 × 3 × q )

= 2 × 3 (p - 2 × q)

= 6 (p - 2q)

(iii) 7a2 + 14a :

let  7a2, 14a can be written as 

 7a2 = 7 × a × a

14a = 2 × 7 × a 

The common factors are given as 2, a

Hence, we get  Algebraic expression as 

 7a2 + 14a = ( 7 × a × a) +( 2 × 7 × a)

= 7 × a (a +2 )

= 7a (a +2)

(iv) – 16 z + 20 z:

let  – 16 z , 20 z3 can be written as 

 – 16 z  = -1 × 2 × 2  × 2  × 2  × z

 20 z3= 2 × 2 × 5 × z × z × z 

The common factors are given as 2, 2, a

Hence, we get  Algebraic expression as 

 – 16 z  +  20 z3= ( -1 × 2 × 2  × 2  × 2  × z) +(2 × 2 × 5 × z × z × z)

= 2 × 2 × z (- 4z +5z2 )

= 4z(- 4z +5z2)

(v) 20 l2 m + 30 a l m :

let 20 l2 m, 30 a l m can be written as 

20 l2 m = 2 × 2 × 5 × I × I × m

30 a l m = 2 × 3 × 5 × a × l × m 

The common factors are given as 2, 5, I, m

Hence, we get Algebraic expression as 

20 l2 + 30 a l m = (  2 × 2 × 5 × I × I × m) +(2 × 3 × 5 × a × l × m )

= (2  × 5  × l × m )[(2 × I) +(3 × a )

= 10Im (2I +3a)  

(vi) 5 x2- 15 xy:

let  5 x2 y , 14a can be written as 

 5 x2 y  = 5 × x × x × y

15 xy2 = 5 × 3 × x × y × y

The common factors are given as 5, x, y

Hence, we get Algebraic expression as 

 5 x2 y  - 15 xy2 = (  5 × x × x × y) -(5 × 3 × x × y × y)

= 5 × x × y(x -3× y )

= 5xy (x -3y)

(vii) 10 a 2 – 15 b2 + 20 c2   

let 10 a 2-15 b2 , 20 c can be written as 

10 a 2 = 2 × 5 × a × a

15 b2 = 5 × 3 × b × b 

20 c2  = 2 × 2 × 5 × c × c

The common factors are given as 5

Hence, we get Algebraic expression as 

10 a 2 – 15 b2 + 20 c2   = ( 2 × 5 × a × a) -( 5 × 3 × b × b ) +(2 × 2 × 5 × c × c)

= 5 ( 2 a 2 – 3 b2 + 4 c)

 (viii) – 4 a2 + 4 ab – 4 ca :

let  – 4 a2  , 4 ab ,– 4 ca can be written as 

– 4 a2  = -× 2 × a × a

4 ab = 2 × 2 × a × b 

– 4 ca =- 2 × 2 × c × a 

The common factors are given as 2, 2, a

Hence, we get Algebraic expression as 

– 4 a2  +4 ab – 4 ca   = ( -× 2 × a × a) +( 2 × 2 × a × b  ) -(- 2 × 2 × c × a )

= 2 × 2 × a  ( -a + b -c)

 (ix) x 2 y z + x y2 z + x y z :

let   x 2 y z ,  x y2 z , x y z can be written as 

 x 2 y z  = x × x × y × z

 x y2 z  = x × y × y × z 

x y z2  = x × y × z × z  

The common factors are given as z, x, y

Hence, we get Algebraic expression as 

 x 2 y z +  x y2 z + x y z = ( x × x × y × z) +( x × y × y × z) + ( x × y × z × z  )

= x × y × z(x + y + z )

= xyz (x + y + z)

(x) a x2 y + b x y2 + c x y z:

let    a x2 y  ,   b x y2 , c x y z can be written as 

 a x2 y  = a × x × x × y

 b x y2   = b × x × y × y

c x y z  = c × x × y × z 

The common factors are given as z, x, y

Hence, we get Algebraic expression as 

 a x2 y  +   b x y2 + c x y z  = ( a × x × x × y) +(  b × x × y × y) + ( c × x × y × z   )

= x × y (ax + by +cz )

= xy (ax + by +cz)

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