Question

15. Factorization of x^2-1-2a-
a^2 is
Select
Search
(x+a+1)(x-a-1)
(x-a+1)(x+a-1)
(x+a+1)(x-a+1)
(x-a-1)(x+a-1)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\: {x}^{2} - 1 - 2a - {a}^{2}$

can be rewritten as

$\rm \: = \: {x}^{2} - (1 + 2a + {a}^{2})$

We know,

$\boxed{ \bf{ {(x + y)}^{2} \: = \: {x}^{2} + {y}^{2} + 2xy }}$

So, using this identity,

$\rm \: = \: {x}^{2} - {(1 + a)}^{2}$

We know,

$\boxed{ \bf{ {x}^{2} - {y}^{2} \: = (x + y)(x - y)}}$

So, using this identity, we get

$\rm \: \: = \:(x - 1 - a)(x + 1 + a)$

Hence,

$\bf :\longmapsto\: {x}^{2} - 1 - 2a - {a}^{2} = (x + 1 + a)(x - 1 - a)$

Option (1) is correct

Additional Information :-

Let us take few more examples!!!

Question :- Factorize the following :-

$\bf :\longmapsto\:(1) \: \: {x}^{2} + 2ax + {a}^{2} - 1$

can be rewritten as

$\rm \: \: = \:( {x}^{2} + 2ax + {a}^{2}) - 1$

$\rm \: \: = \: {(x + a)}^{2} - 1$

$\rm \: \: = \: {(x + a)}^{2} - {1}^{2}$

$\rm \: \: = \:(x + a + 1)(x + a - 1)$

Hence,

$\bf :\longmapsto\:\: \: {x}^{2} + 2ax + {a}^{2} - 1 = (x + a + 1)(x + a - 1)$

$\bf :\longmapsto\:(2) \: \: {x}^{4} - {(y - z)}^{4}$

$\rm \: \: = \: {( {x}^{2} )}^{2} - {\bigg( {(y - z)}^{2} \bigg) }^{2}$

$\rm \: \: = \:\bigg( {x}^{2} + {(y - z)}^{2} \bigg)\bigg( {x}^{2} - {(y - z)}^{2} \bigg)$

$\rm \: \: = \:\bigg( {x}^{2} + {(y - z)}^{2} \bigg)\bigg( ({x} - {(y - z)})(x + y - z) \bigg)$

$\rm \: \: = \:\bigg( {x}^{2} + {(y - z)}^{2} \bigg)\bigg( ({x} - y + z))(x + y - z) \bigg)$