Question

Factorise 64a square − (7c + 4b) the whole square

Answer 1

Step-by-step explanation:

Solution : 64a² - 4b² - c² + 4bc

= 64a² - ( 4b² + c² - 4bc )

= 64a² - [ (2b)² + c² - 2 × 2b × c ]

Since a² + b² - 2ab = ( a - b )²

= 64a² - [( 2b - c )²]

= (8a)² - [( 2b - c )²]

Since a² - b² = ( a + b )( a - b )

= [ 8a + ( 2b - c )][ 8a - ( 2b - c )]

= ( 8a + 2b - c )( 8a - 2b + c )

Answer 2

Answer:

(8a - 7c + 4b)(8a + 7c + 4b)

Step-by-step explanation:

64a² - (7c + 4b)² (by a² - b² = (a+b) (a-b))

(8a)² - (7c + 4b)²

(8a - 7c + 4b)(8a + 7c + 4b)