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 .