Factorizing using Product of Two Binomials Identity

Product of Two Binomials Identity:

( x + a)( x + b ) = x² + (a + b) x + ab. 

Factorize the following:

 1) p² + 10pq + 21   

Answer:

 In the given expression : p² q² + 10pq + 21, the number of terms are three.

1^st term is a Perfect square, where as the 3^rd term is not a Perfect square. Here,we have to use Product of Two Binomials Identity.

x² + (a + b) x + ab = ( x + a)( x + b ) .

In this , we need to find  common factors : N_1,  N₂

p² q²  * 21 = 21 p² q² ,the multipliers of  21 p² q²  are 7pq and 3 pq .

Therefore, the factors  N₁ ,  N_{2 =}  7 pq , 3 pq.

The middle term is N₁  + N₂ =  7pq + 3pq = 10 pq.

So, the expression :  p² q² + 10pq + 21   is written as in terms of their product of factors.

 = p² q² + 7pq + 3pq + 21

 =  ( pq + 7 ) ( pq + 3).     

                            (OR)

 p² q² + 10pq + 21    

Answer:

The expression is not a perfect square.To make it into Perfect square expression ,we need to add 4  and subtract 4 to keep the expression unchanged .

Therefore,the expression changes into:

 p² q² + 10pq + 21 + 4 – 4

=  (p² q² + 10pq + 25 ) – 4

= (  (pq)² + 2 * pq * 5 + (5)² ) – 4    

( (pq)² + 2 * pq * 5 + (5)² ;it is in the form of  Squares of Binomials Identity: 

a² + 2 ab + b² = (a+b)²  ;  Therefore,

= ( pq + 5 )² – 2²     

It is in the form of  Differrence of Squares Identity :

 a²  - b² = ( a+b)(a - b)

Here, a = pq+5  and  b = 2.

Therefore; ( pq + 5 )² – 2²  =  ( pq +5 +2 ) ( pq + 5 – 2 )    

                                              = ( pq + 7 ) ( pq + 3 ).

 

         

2) u² + uw + uv + vw

Answer:

The expression: u² + uw + uv + vw  is given in terms of their factors.

By using the Product of Two Binomials Identity: 

x² + (a + b) x + ab = ( x + a)( x + b ) .

 we need to factorize this expression.

The common Factor is taken out from the expression: u is common in first two terms and v is common in last two terms.Therefore;

= u² + uw + uv + vw

= u ( u + w) + v( u + w )

= (u + w ) ( u + v) .