It's true up to a constant that over an algebraically closed field all polynomials factor in the way you describe. The correct statement is that every polynomial factors as
$$f(x) = b(x-a_1)\dots(x-a_n)$$ for $a_i,b \in K$. The point is you can't guarantee the leading coefficient is $1$ (consider the polynomial $2x-1$ for instance). Note also, that we could have $a_i = a_j$ for some $i \neq j$.
The best way to phrase this is probably to say that every polynomial factors as a product of linear polynomials or splits into linear factors.
The roots refer to the elements $a_i \in K$ such that $f(a_i)=0$. While it's true that these (counted with multiplicities) determine the factorization of $f$ up to a constant, these do not refer to the linear factors themselves.