25a²-4b²-28bc-49c². Factorise the equation step by step.
Answers
Answered by
0
Answer:
25 a² -( (2b)²+ (7c)² -2* 7*2*b*c)
(5a)²- (2b - 7c)²
(5a + 2b-7c)( 5a - 2b+7 c)
Answered by
0
Solution:
25a^2 - 4b^2 - 28bc - 49c^2
Since both terms are perfect squares, factor using the difference of squares formula
a^2 - b^2 = (a + b) (a - b)
where
a = 5a
b = 2b + 7c
Therefore, the factor of 25a^2 - 4b^2 - 28bc - 49c^2 is
(5a + 2b + 7c) (5a - 2b - 7c)
Hope this will be helpful to you.
25a^2 - 4b^2 - 28bc - 49c^2
Since both terms are perfect squares, factor using the difference of squares formula
a^2 - b^2 = (a + b) (a - b)
where
a = 5a
b = 2b + 7c
Therefore, the factor of 25a^2 - 4b^2 - 28bc - 49c^2 is
(5a + 2b + 7c) (5a - 2b - 7c)
Hope this will be helpful to you.
Similar questions