Advertisements
Advertisements
प्रश्न
Factorise : (a2 + b2 - 4c2)2 - 4a2b2
Advertisements
उत्तर
(a2 + b2 - 4c2)2 - 4a2b2
= ( a2 + b2 - 4c2 )2 - ( 2ab )2
= ( a2 + b2 - 4c2 - 2ab )( a2 + b2 - 4c2 + 2ab ) [ ∵ a2 - b2 = ( a + b )( a - b )]
= ( a2 + b2 - 2ab - 4c2 )( a2 + b2 + 2ab - 4c2 )
= [ ( a - b )2 - ( 2c )2 ][ ( a + b )2 - ( 2c )2]
= ( a - b + 2c )( a - b - 2c )( a + b + 2c )( a + b - 2c )
APPEARS IN
संबंधित प्रश्न
Factorise : a2 - 81 (b-c)2
Factorise : a4 - 1
Factorise:
(a2 − 1) (b2 − 1) + 4ab
Factorise : `x^2 + 1/x^2 - 11`
Factorise : `4x^2 + 1/(4x)^2 + 1`
Factorise the following by the difference of two squares:
441 - 81y2
Factorise the following by the difference of two squares:
`"m"^2 - (1)/(9)"n"^2`
Factorise the following:
4xy - x2 - 4y2 + z2
Factorise the following:
4x2 - 12ax - y2 - z2 - 2yz + 9a2
Express each of the following as the difference of two squares:
(x2 - 2x + 3)(x2 + 2x + 3)
