Question

factorise : 25a2 - 4b2 + 28bc - 4ac².​

Answer 1

/* There is a mistake in the question */

It must be like this .

$25a^{2} - 4b^{2} + 28bc - 49c^{2}$

$= (5a)^{2} - [ 4b^{2} - 28bc + 49c^{2}]$

$= (5a)^{2} - [ (2b)^{2} - 2\times (2b) \times (7c) + (7c)^{2} \\= (5a)^{2} - ( 2b - 7c)^{2}$

By Algebraic Identity :

$\boxed { \pink { x^{2} - 2xy + y^{2} = ( x - y )^{2} }}$

$= [ 5a + ( 2b - 7c )] [ 5a - ( 2b - 7c )]$

By Algebraic Identity :

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

$= ( 5a + 2b - 7c )( 5a - 2b + 7c )$

Therefore.,

$\red { 25a^{2} - 4b^{2} + 28bc - 49c^{2} }$

$\green { = ( 5a + 2b - 7c )( 5a - 2b + 7c )}$

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