Advertisements
Advertisements
Question
Factorise : 4a2 - (4b2 + 4bc + c2)
Sum
Solution
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 ]
shaalaa.com
Method of Factorisation : Difference of Two Squares
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
Factorise : a2 - (2a + 3b)2
Factorise : a2 - 81 (b-c)2
Factorise : 50a3 - 2a
Factorise : 4a2b - 9b3
Factorise : 3a5 - 108a3
Factorise : a3 + 2a2 - a - 2
Factorise : (a + b)3 - a - b
Factorise : 4xy - x2 - 4y2 + z2
Factorize : 9a2 - (a2 - 4) 2
Factorise the following by the difference of two squares:
x2 + y2 - z2 - 2xy
