Question

Factorise 1−a2−b2−2ab

Answer 1

1-a²-b²-2ab
=1-(a²+b²+2ab)
=1- (a+b)²

=1²- (a+b)²
=[1+ (a+b)][1- (a+b)]
=[1+a+b][1-a-b]

Answer 2

Given:

• An expression $1-a^2-b^2-2ab$ is given

To Find:

• Factories the given equation.

Solution:

Given equation  is

                      ⇒  $1-a^2-b^2-2ab$

We can re-write it as

                     ⇒  $1 - (a^2 + b^2 - 2ab)$

Using the identity

                     ⇒   $a^2+ b^2 - 2ab = (a - b)^2$

Putting above value, we get

                     ⇒   $1 - ( a - b )^2$

                     ⇒    $(1)^2 - ( a - b )^2$

Again using the identity -

                   ⇒     $a^2 - b^2 = ( a + b ) ( a - b )$

By using above identity

we get,

                    ⇒ $( 1 + a - b )( 1 - a + b )$

Thus, the factors are $( 1 + a - b ) and ( 1 - a + b )$