Question

Factorize the given expression  $ 1 + 2ab - ({a^2} + {b^2}) $.

Answer 1

Hint: The given problem is an expression containing two variables  $ a $  and  $ b $ and is of degree  $ 2 $  .We will use the binomial formula to the given problem and try to add and cancel the terms wherever necessary to obtain a factorized form of the given equation.

The binomial formula is used to express two added or subtracted terms raised to some natural numbers  $ n $  in the form of individual terms raised to some natural number less than or equal to  $ n $.

Complete step-by-step answer:

The binomial formula for power two is:

 $ {(a - b)^2} = {a^2} - 2ab + {b^2} $  

And the product of  $ (a + b) $  and  $ (a - b) $  is given by:

 $ (a + b)(a - b) = {a^2} - {b^2} $ 

The given expression is:

$ 1 + 2ab - ({a^2} + {b^2}) $ 

Adding and subtracting  2ab

$  = 1 + 2ab - ({a^2} + {b^2}) + 2ab - 2ab $      

Take the  +2ab  term inside the bracket to make the factorization easier. Note that this manipulation is the most important part of solving this problem.

 $  = 1 + 2ab - ({a^2} + {b^2} - 2ab) - 2ab $ 

 $  = 1 + 2ab - ({a^2} - 2ab + {b^2}) - 2ab $ 

 $  = 1 + 2ab - {(a - b)^2} - 2ab $ 

 $  = 1 + (2ab - 2ab) - {(a - b)^2} $ 

Cancel the terms in the bracket and rewrite the expression by simplifying it

$  = 1 - {(a - b)^2} $ 

$ 1 $  raised to any power is equal to  $ 1 $  or  $ {1^n} = 1 $  , so we can write \[{1^2} = 1\]. So, the above expression is same as:

$ \Rightarrow {1^2} - {(a - b)^2} = [1 + (a - b)][1 - (a - b)] $ 

$  = (1 + a - b)(1 - a + b) $ 

Therefore, the factorization of given expression  $ 1 + 2ab - ({a^2} + {b^2}) $ is $  = (1 + a - b)(1 - a + b) $ .

Note: In the intermediate steps it is necessary to check if there is a chance to cancel terms which will make simplification easy. If all terms are chosen randomly then factorization will not be possible. Make special attention towards the change in signs as:

a) Multiplication of two negative terms is negative.

b) Multiplication of two positive terms is positive.

c) Multiplication of a negative and a positive term is negative.