Question

Factorise: 4a² - ( 4b² + 4bc + c²)​

Answer 1

Answer:

4a2 - (4b2 + 4bc + c2)

= ( 2a )2 - ( 2b + c )2

= [ 2a - ( 2b + c )][ 2a + (2b + c )] [ ∵ a2 - b2 = ( a + b )( a - b )]

= [ 2a - 2b - c ][ 2a + 2b + c ]

Step-by-step explanation:

hope it may helps u...

Answer 2

$4a² - ( 4b² + 4bc + c²)$

${(2a)}^{2} - (2b + c)^{2}$

Difference of two squares-->

$(2a - 2b - c)(2a + 2b + c)$

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Here why I changed

$( 4b² + 4bc + c²)$

This to

${(2b + c)}^{2}$

Reason::

Here I used this

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

And it was forming the same

${(2b + c)}^{2} = 4 {b}^{2} + {c}^{2} + 4bc$

So, that's why....