25x^2-(4y^2+49z^2-28yz)?

Answer 1

25x^2 - (4y^2 + 49z^2 - 28yz)
first you should rearrange inside the parenthesis to better see the factors
25x^2 - (4y^2 - 28yz + 49z^2)
now you can see that 4y^2 = (2y)(2y) and 49z^2 = (7z)(7z) so you should write them as factors of the part in parenthesis
25x^2 - (2y?7z)(2y?7z) what sign should the ? be changed to?

+ * + = +
+ * - = -
- * + = -
- * - = + <-- this is the one you want because you have a + on the last term and a - for the next to last and this will give you that
25x^2 - (2y-7z)^2
now you have the "Difference of Perfect Squares" so you can factor this...

(5x - (2y-7z))*(5x + (2y-7z))

Answer 2

25x^2-(4y^2+49z^2-28yz)
25x^2 - (2y - 7z) (2y - 7z)

Answer 3

25x^2-(4y^2+49z^2-28yz)?
= 25x^2 - (2y-7z)^2
= (5x +2y-7z)(5x -2y+7z)