Finding Big-O is pretty straightforward for an algorithm where $f(n)$ is
$$f(n) = 3n^4 + 6n^3 + 10n^2 + 5n + 4$$
The lower powers of $n$ simply fall off because in the long run $3n^4$ outpaces all of them. And with $g(n) = 3n^4$ we say $f(n)$ is $O(n^4)$.
But what would Big-O be if instead of 3 we were given a really small constant, for example $$f(n) = 0.0000000001n^4 + 6n^3 + 10n^2 + 5n + 4$$
Would we still say $f(n)$ is $O(n^4)$?