why O(2n^2) and O(100 n^2) same as O(n^2) in algorithm complexity?

why O(2n^2) and O(100 n^2) same as O(n^2) in algorithm complexity?

I am new in algorithm analysis domail I read here in this post Plain
English explanation of Big O because if n = 4 then NO OF OPERATIONS WILL
BE O (2 n^2) = 32 operations O(100 n^2) = 1600 operations and O (n^2) = 16
operations
can any one can explain this ?