Factorizing using Difference of Squares Identity

 Difference of Squares Identity:

  a² – b² = (a – b )(a + b)

  

Factorization using Difference of squares Identity

1) 16x² – 36y²

Answer:

In the given expression : 16x² – 36y²

There are two variables: x , y . The expression has two terms,which are Perfect squares and the sign is Negative. 

1^st term , a² =  16 x²  ; a = 4x

2^nd term, b² = 36y²  ; b = 6y                  

Here, the middle term is not their ,

By using the Difference of Squares Identity: a² – b² = (a – b )(a + b)

Therefore, factorization of 16x² – 36y²

                                       = (4x)²  - (6y)²    

                                       = ( 4x + 6y ) ( 4x – 6y).

 

2) 1 – 25(2a – 5b)²

Answer:

In the given expression: 1 – 25(2a – 5b)²

The number of terms = 2 ,which are perrfect squares and the sign is Negative.

1^st term: a² = 1 ; a = 1

2^nd term: b² =  25(2a – 5b)² ;

              b = (5)² (2a – 5b)² = ( 5( 2a – 5b))² = ( 10a – 25b )²

Here,also Middle term is not their.

By using the Difference of Squares Identity: a² – b² = (a + b) (a – b )

Therefore, factorization of :  1 – 25(2a – 5b)²  = 1² – ( 10a – 25b)²

                                       =  ( 1 + 10a – 25b )( 1 – (10a - 25b))

                                        = ( 1 + 10a – 25b) ( 1 – 10a + 25b).

 

3) m²/n² - 36        

Answer:

The given expression: m²/n² - 36 , is a perfect square expression and has two terms,no middle term ; with Negative sign.

1^st term: a² = m²/n²  ; a = m/n  

2^nd term: b² = 36 ; b = 6

The given expression: m²/n² – 36 ; is in the form of :

a² – b² = (a + b) (a – b ).

Therefore , factorization of : m²/n² – 36

                                            =  (m/n)² - 6²

                                         = ( m/n + 6 )(m/n – 6 ).  

4) (x – 2)² – ( x – 3)²

Answer:

The given expression: (x – 2)² – ( x – 3)² ; is a perfect square form and it is in the form of : a² – b² = (a + b) (a – b ).

Here, a² =  (x – 2 )² ; a = (x – 2 )

          b² = (x – 3 )² ; b = ( x – 3 )

Therefore, factorization for: (x – 2)² – ( x – 3)²

 = ( (x–2) + ( x–3) ) ( (x - 2) – (x – 3) ; solving the terms in brackets

 = ( x -2 + x – 3 ) ( x- 2 – x + 3)

 = ( 2x -5) ( 1)

 = ( 2x – 5 ).                                     

 

5) 25(x+y)² – 36(x- 2y)²                                            

Answer:

The given expression: 25(x+y)² – 36(x- 2y)² , is a perfect square form.The expression is written as : 5²(x+y)² – 6²(x -2y)² = ( 5(x+y))² – ( 6(x – 2y))² ; which is in the form of:  : a² – b² = (a + b) (a – b ).

Here, a² =  (5(x+y))² ; a = 5(x+y) = 5x + 5y

          b² = (6(x-2y))² ; b = 6(x-2y) = 6x – 12y

Therefore, factorization of :  25(x+y)² – 36(x- 2y)²

 = (( 5x+5y) + (6x-12y)) (( 5x+5y) - (6x-12y))    

 = ( 5x+5y+6x-12y) (5x+5y – 6x + 12y)

 = ( 11x – 7y) ( -x + 17y)

 = ( 11x – 7y) ( 17y – x ).

 

6) x³ – 25x

Answer:

We need to change the given expression in to Perfect square form.

Here, We take ‘x’ as common from the expression : x³ – 25x , as it change into Perfect square. Therefore, x³ – 25x = x( x² – 25).

The new expression is: x( x² – 25) , it is again written as: x( x² – 5² ).

The expression in the brackets, i.e: ( x² – 5² ) is in the form of :

 a² – b² = (a + b) (a – b ).

Here, a² = x² ; a = x

          b² = 5² ; b = 5

Therefore , factorization of : x³ – 25x

= x( x² – 5² )

= x ( x+ 5 ) ( x – 5 ).

 

7) 3t² – 27g²

Answer:

Here, also we have to take 3 as common from the expression 3t² – 27g² , to make it into Perfect square form.

The new expression is : 3 ( t² – 9g² )

  = 3 (t² – (3g)² )  ;  since 9g² = 3g * 3g = (3g)²    

Hence,the expression: (t² – (3g)² ); is in the form of :

 a² – b² = (a + b) (a – b ).

Here, a² = t² ; a = t

          b² = 3g² ; b = 3g

Therefore, factorization of : 3t² – 27g²

= 3 (t² – (3g)² ) 

= 3 ( t + 3g) ( t – 3g).