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). 

Factorising using Squares of Binomial Identity

Square of a Binomial:

i) (a + b)² = a² + 2 _* a _* b +  b² = a² + 2 ab + b² 

                  = (a + b)( a + b) .

                           (or)

ii)    (- a – b )² =  a² + 2 _* a _* b + b^2. =  a²  + 2 ab + b²   

                       =  (a + b) ( a + b).

iii)  (a – b)²  =  a² – 2_* a _* b + b²  =  a² – 2ab + b²  

                      =  ( a – b ) ( a – b ).    

Note : The quadratic equation or binomial  ax² + bx + c is a perfect square ; if b² = 4ac.

Factorize the expressions using Identities:

1)      4x² + 12xy + 9y²

Answer:

In the given expression 4x² + 12xy + 9y² 

There are two variables: x , y and has three terms.1^st , 3 ^rd  terms are  Perfect squares.

The middle term has Positive sign.

1^st  term : a² = 4x²  ; a = 2x

3 ^rd  term : b² = 9y²  ; b = 3y

Middle term (12xy)  = 2  _*  1^st term  _* 3 ^rd term = 2 _* a _* b

                      = 2 _* 2x _* 3y   = 12xy

By using the Square of Binomial Identity : 

(a + b )²  =  a² + 2 ab + b²  =  (a + b) ( a + b)

Therefore, factorization of : 4x² + 12xy + 9y²

                                        = ( 2x + 3y ) ²

                                         = (2x + 3y) ( 2x + 3y).      

  

 2)      4x² – 4x + 1

Answer:

In the given expression  4x² – 4x + 1

There are two variables: x , y and has three terms.1^st , 3 ^rd  terms are  Perfect squares.

The middle term has Negative sign. 

1^st term,  a² = 4x²  ; a = 2x

3^rd term, b² = 1; b = 1.

Middle term, (- 4x ) =  - 2 _* 2x _* 1  =  - 4x  = 2 ab

By using the Square of Binomial Identity :

(a -  b )²  =  a²  -  2 ab  + b² = ( a – b)( a – b)

Therefore, factorization of 4x² – 4x + 1

      = ( 2x – 1 )²   

      =  ( 2x - 1 ) ( 2x - 1).

 

3 ) w² – 2w/x + 1/ x²

Answer:                                                   

The given expression is written as : w² – 2w/x + 1/ x² ;also 1^st , 3 ^rd  terms are  Perfect squares . The middle term has Negative sign. The expression is written as:

= w² – 2 * w * 1/x + (1/x)² ; it is in the form of :

(a -  b )²  =  a²  -  2 ab  + b² = ( a – b)( a – b)

Here, a = w , b = 1/x , ab = w/x ; 2 ab = 2w/x.

Therefore ; factorization of :w² – 2w/x + 1/ x²

= ( w – 1/x)²

= ( w – 1/x)( w - 1/x) .

 

4) 1/t² – 8/t + 16

 Answer:

The given expression is written as: 1/t² – 8/t + 16 ;also 1^st , 3 ^rd  terms are  Perfect squares . The middle term has Negative sign. The expression is written as:

= 1/t² – 2 * 1/t * 4 + 4²  ; it is in the form of :

(a -  b )²  =  a²  -  2 ab  + b² = ( a – b)( a – b)

Here, a = 1/t ; b = 4 ; ab = 4/t ; 2 ab = 8/t .

Therefore ; factorization of : 1/t² – 8/t + 16

= ( 1/t – 4 )²

= (1/t – 4)( 1/t – 4).

 

5) p² + 169 +26p   

Answer:

The given expression is written as: p² + 169 +26p ;rearranged as : p² + 26p + 169     

In the given expression :1^st , 3 ^rd  terms are  Perfect squares . The middle term has Positive sign. The expression is written as:

= p² + 2 *p * 13 + 13² ; it is in the form of :

(a + b )²  =  a² + 2 ab + b²  =  (a + b) ( a + b)

Here, a = p ; b = 13 ; ab = 13p ; 2ab = 26p.

Therefore; factorization of : p² + 169 +26p   

= ( p + 13 )²

= ( p + 13 ) ( p + 13 ).

6) (3a – 5b)² + 2 (3a – 5b)(2b –a) + ( 2b – a )²          

Answer:

In this expression : (3a – 5b)² + 2 (3a – 5b)(2b –a) + ( 2b – a )²        

The number of terms are Three ,1^st and 3^rd terms are Perfect squares . The middle term has Positive sign.The expression is in the form of : a² + 2 ab + b² 

1^st term, a² =  ( 3a – 5b )² ; a = (3a – 5b)

3^rd term, b² = ( 2b – a )² ; b =  ( 2b – a )

Middle term : 2 ab =  2 (3a – 5b)(2b –a)

By using the Square of Binomial Identity : 

(a + b )²  =  a² + 2 ab + b²  = ( a + b )( a + b )

Solving : ( a + b ) = (( 3a – 5b ) + ( 2b – a )) ,

                             = ( 3a – 5b + 2b – a ) = (2a – 3b )

Therefore, factorization of : (3a – 5b)² + 2 (3a – 5b)(2b –a) + ( 2b – a )²

                                        =  ( ( 3a – 5b ) + ( 2b – a ) )²  ;it is in the form:  ( a + b)²  

                                  =  ( 2a – 3b )²    ;   since : ( a + b = 2a - 3b )

                              =  (2a – 3b ) ( 2a – 3b) .     

Factorizing using Cubes of Trinomial Identity

 CUBES OF TRINOMIAL IDENTITY :

 a³ + b³ + c³ = ( a + b + c )( a² + b² + c² – ab – bc – ca ) +  3abc  .

 If  ( a + b + c ) = 0 ; then a³ + b³ + c³ = 3 abc .

 Factorize the expressio using Algebraic Identities:

1)  ( a - b)³ + ( b – c )³ + ( c – a )³

Answer:

The given expression: ( a - b)³ + ( b – c )³ + ( c – a )³ ; is in the form of  : a³ + b³ + c³

Here;  a = a – b ; b = b – c ; c = c – a .

By using the Cubes of Trinomial Identity :

a³ + b³ + c³ = ( a + b + c )( a² + b² + c² – ab – bc – ca ) +  3abc ;

 If a + b + c = 0 ,then  a³ + b³ + c³ = 3 abc

substituting  a, b , c values in the above Identity we get:

a + b + c = a – b +  b – c + c – a  = 0; 

( a - b)³ + ( b – c )³ + ( c – a )³

=  0 + 3 ( a-b)*( b-c) * (c-a)

= 3 (a-b)(b-c)(c-a). 

2)  x³ + y³ - 3xy +1.

Answer:

The given expression : x³ + y³ - 3xy +1 , can be re-write as;

= x³ + y³ + 1³ – 3* x * y * 1 ; which is in the form of

a³ + b³ + c³ – 3abc = ( a + b + c )( a² + b² + c² – ab – bc – ca )

Here, a =x ; b = y ; c = 1 ; a²  = x²  ; b² = y² ; c² = 1 ; ab= xy ; bc = y ; ca = x

Therefore ;  x³ + y³ + 1 – 3 xy

= ( x + y + 1 ) (x² + y² + 1 – xy – y – x )

= ( x + y + 1 ) (x² + y² – xy –y – x + 1).

 

3) 8x³ + 27y³ – 90xy + 125

Answer:

The given expression : 8x³ + 27y³ – 90xy + 125 , can be rewrite as :

=   8x³ + 27y³ + 125 – 90xy

=  ( 2x)³  + ( 3y)³ + 5³  - 3 * 2x * 3y * 5 ; which is in the form of

a³ + b³ + c³ – 3abc = ( a + b + c )( a² + b² + c² – ab – bc – ca )

Here, a = 2x ; b = 3y ; c= 5 ; a² = 4x² ; b² = 9y² ; c² = 25

ab = 6xy ; bc = 15y ; ca = 10x.

Therefore ;  8x³ + 27y³ + 125 – 90xy

= ( 2x + 3y + 5 )( 4x² + 9y² + 25 – 6xy – 15y – 10x)

= ( 2x + 3y + 5 )( 4x² + 9y²  - 6xy – 15y – 10x + 25 ).

Factorization using Squares of Trinomial Identities.

 Squares of Trinomial Identities :

 i) ( a + b + c )²  = a² + b² + c² + 2 ab + 2 bc + 2 ca .

ii) ( a + b – c )²   = a² + b² + c² + 2 ab  - 2 bc – 2 ca .

iii) ( a – b – c )²  =  a² + b² + c²  - 2 ab + 2 bc – 2 ac.

iv) ( - a + b + c )² =  a² + b² + c² -  2ab  + 2 bc – 2 ac.

v) (  a – b  +  c )²  =  a² + b² + c² -  2ab  - 2 bc + 2 ac.

1. Factorize:  2x²  + y²  + 8z² - 2 √2 xy + 4 √2 yz – 8zx

Answer:

In the given expression 2x²  + y²  + 8z² - 2 √2 xy + 4 √2 yz – 8zx

The number of variables are three : x , y ,z

By using the Square of a Trinomial Identity :

(i) (a – b – c )² =   a² + (-b)² + ( -c)²  + 2 * a * (-b) + 2 * ( -b) * (-c) + 2 * a * (-c)

                   =  a² + b² + c²  - 2 ab + 2 bc – 2 ac.    

                            (or)

(ii) ( - a + b + c )²  =  a² + b² + c² -  2ab  + 2 bc – 2 ac.

Here, we can use either of the identity. When we use this identity

( a – b – c )²  =  a² + b² + c²  - 2 ab + 2 bc – 2 ac , the result is:

1^st term   : a² = 2x²    ;     a = √2  x

2 ^nd term : b²  =  - 1 y² ; b =  - y

3^rd term : c² =  - 8z²  ;   c = - 2√2  z

4^th term : 2ab = 2 * (√2  x) * (-y) =  - 2 √2  xy

5^th term : 2bc = 2 * ( - y ) * (- 2√2  z ) =  4√2  yz

6^th term : 2ac = 2 *  √2  x * (- 2√2  z ) =  - 8zx

Therefore , the expression  :  

2x²  + y²  + 8z² - 2√2  xy + 4√2  yz – 8zx  is written as:

= (√2  x  - y - 2√2  z )²

= (√2  x  - y - 2√2  z)(√2  x  - y - 2√2  z)

    

                                 (OR)

The  expression  2x²  + y²  + 8z² - 2 √2 xy + 4 √2 yz – 8zx  is factorized in other way, when we use this identity : (- a + b + c )²  =  a² + b² + c² -  2ab  + 2 bc – 2 ac , to Factorize the given expression:  2x²  + y²  + 8z² - 2 √2 xy + 4 √2 yz – 8zx , the result is:

1^st term   : a² =  -2x²    ;     a =  - √2  x

2 ^nd term : b²  =  1 y² ; b =   y

3^rd term : c² =  8z²  ;   c =  2√2  z

4^th term : 2ab = 2 * (-√2  x) * (y) =  - 2 √2  xy

5^th term : 2bc = 2 * (  y ) * ( 2√2  z ) =  4√2  yz

6^th term : 2ac = 2 * ( - √2  x) * ( 2√2  z ) =  - 8zx

Therefore , the expression  :  

2x²  + y²  + 8z² - 2√2  xy + 4√2  yz – 8zx , is factorized as:

 = ( -√2  x  + y + 2√2  z )²

= ( - √2  x + y + 2√2  z)( - √2  x + y + 2√2  z).

2) k² - 18k + 1/k² – 18/ k + 83.

Answer:

The given expression: k² - 18k + 1/k² – 18/ k + 83 is rearranged as: keeping the square terms at one side; k² + 1/k² – 18k – 18/ k + 83.

= k² + 1/k² – 18k – 18/ k + 81 + 2     ( as we can write  83 = 81 + 2 )         

= k² + 1/k² + 81 – 18k – 18/k + 2

= k² + 1/k² + 9² – (2 * k * 9 ) – (2 * 9 * 1/k) +( 2 * k * 1/k) ; rearranged as:

= k² + 1/k² + 9² + ( 2 * k * 1/k) – (2 * 1/k * 9) – (2 * k * 9 )

By seeing the above expression ; 5^th term and 6^th term has Negative signs. So, we can say that the

3^rd term has negative sign in the expression. Therefore;

= k² + 1/k² + (- 9 )² + ( 2 * k * 1/k) – (2 * 1/k * 9) – (2 * k * 9 ) ; which is in the form of Square of Trinomials Identity:  ( a + b – c )²  = a² + b² + c² + 2 ab  - 2 bc – 2 ca .

Here, a = k ; b = 1/k ; c= -9 ; 2 ab = 2 ; 2 bc = -18 / k ; 2 ca = -18k .

= ( k + 1/k -9)²

= (k + 1/k -9) (k + 1/k -9) . 

3) m² + n² -2mn – 6n  + 6m + 9

Answer:

The expression m² + n² -2mn – 6n  + 6m + 9 , is written as:

 m² + n² -2mn – 6n  + 6m + 3² , writing the squared terms at one place and rearranging the expression.   

=  m² + n² + 3² – 2mn - 6n + 6m ;

 Here, the 4^th term and 5^th term has Negative sign.By this we can say 2^nd term must be Negative term .The expression  m² + n² + 3² – 2mn - 6n + 6m ; is written as:

= m² + (- n )² + 3² + ( 2 * m * ( -n))  + (2 * (-n) * 3) + ( 2 * m * 3)    

which is in the form of  Square of Trinomial :

 ( a – b  + c )²  =  a² + b² + c²  - 2 ab  -  2 bc + 2 ac.

Here, a = m ; b = -n ; c = 3 ; 2ab = - 2mn ; 2 bc = - 6n ; 2 ca = 6m .

Therefore: m² + n² + 3² – 2mn - 6n + 6m = ( m – n + 3 )²

= ( m – n + 3 ) ( m – n + 3 ).

 

4)  x² + y² + 2xy + 1/ z² + 2y / z + 2x / z

Answer: Re- arranging the expression :

x² + y²  + 1/ z² + 2xy + 2y / z + 2x / z .

Here , a= x ; b= y ; c = 1 / z

2 ab = 2 * x * y = 2 xy

2 bc = 2 * y * 1 / z = 2y / z

2 ca = 2 * 1/z * x = 2x / z

By using the Square of Trinomial Identity: ( a + b + c )² = a² + b² + c² + 2 ab + 2 bc + 2 ca

The given expression x² + y² + 2xy + 1/ z² + 2y / z + 2x / z , is written as

= ( x + y + 1 /z )²

= ( x + y + 1/z )( x + y + 1/z ).    

5)  ( 3x – 2y – z )²

Answer:

The given expression ( 3x – 2y – z )² ,is in the form of  : ( a – b – c )²

We know that : ( a – b – c )² = a² + b² + c²  - 2 ab + 2 bc  -  2 ca .

Here,   a = 3x ; b = (-2y) ; c= (-z)

Therefore,  ( 3x - 2y – z )² = ( 3x + ( -2y) + ( -z) )²

 = (  (3x)² + ( -2y)² + ( -z)² + 2* 3x * ( -2y) +  2 * ( -2y) * ( -z) + 2 * ( -z) * ( 3x) )

=  9x² + 4y²  + z²  - 12xy + 4yz – 6zx .

 

6) Simplify a+ b+c =25 and ab + bc + ca = 59. Find the value of  a² + b² + c²  ?

Answer:

The data given in the Problem is : a + b + c = 25 ;

Squaring the equation both sides, we get:

 (a + b + c)²  = 25²   ; L.H.S is in the form of Square of Trinomial.

We know : (a + b + c)²  =   a² + b² + c²  + 2 ab + 2 bc  + 2 ca

 a² + b² + c²  + 2 ab + 2 bc  + 2 ca  =  625   ; (expanding L.H.S)

 a² + b² + c²  + 2 ( ab + bc + ca ) = 625        ; ( taken 2 as common in L.H.S )

 a² + b² + c²  + 2 ( 59 ) = 625         ;        (  Substituting:   ab + bc + ca = 59 )

 a² + b² + c²  +  118 = 625

 a² + b² + c²   =  625 – 118

Therefore ,  a² + b² + c²   =  507 .