Question

Factorise
(X^2+4)-2a-a^2-5

Answer 1

Answer:

(x+a+1) (x-a+1)

Step-by-step explanation:

(x^2+4) -2a-a^2-5

• opening the bracket x^2+4-2a-a^2-5
• x^2-a^2-2a-1
• taking x^2 ans - common
• x^2-(a^2+2a+1)
• x^2-(a+1)^2
• (x)^2-(a+1)^2
• as they are in a^2 - b^2 form
• so then ans is (x+a+1) ( x-a+1)...

I think this will help u solve it

Answer 2

( x² + 4 ) - 2a - a² - 5 = (x + a + 1)(x - a - 1)

Given :

The expression ( x² + 4 ) - 2a - a² - 5

To find :

To factorise the expression

Formula :

• (a + b)² = a² + 2ab + b²

• a² - b² = ( a + b ) ( a - b )

Solution :

Step 1 of 2 :

Write down the given expression

The given expression is

( x² + 4 ) - 2a - a² - 5

Step 2 of 2 :

Factorise the expression

$\sf ( {x}^{2} + 4) - 2a - {a}^{2} - 5$

$\sf = {x}^{2} + 4 - 2a - {a}^{2} - 5$

$\sf = {x}^{2} - 2a - {a}^{2} - 1$

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

$\sf = {x}^{2} - ({a}^{2} + 2.a .1+ {1}^{2} )$

$\sf = {x}^{2} - {(a + 1)} ^{2} \: \: \bigg[\: \because \: {a}^{2} + 2ab + {b}^{2} = {(a + b)}^{2} \bigg]$

$\sf = (x + a + 1)(x - a - 1)\: \: \bigg[\: \because \: {a}^{2} - {b}^{2} = (a + b)(a - b) \bigg]$

∴ ( x² + 4 ) - 2a - a² - 5 = (x + a + 1)(x - a - 1)

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