We show that $h$ must be in $R[x]$. Suppose that we have polynomials
$$g = x^n + a_1x^{n-1} + ... + a_n \in R[x]$$ $$h = x^m +b_1x^{m-1} + ... + b_m \in K[x]$$ such that $f = gh \in R[x]$ but $h \notin R[x]$. Let $r:= \min {i | \{i \mid b_i \notin R}$R\}$. Since$f \in R[x]$we have$b_r + a_1b_{r-1} + ... + a_{r-1}b_1 + a_r \in R$, so$b_r \in R$. It is a contradiction. 1 We show that$h$must be in$R[x]$. Suppose that we have polynomials $$g = x^n + a_1x^{n-1} + ... + a_n \in R[x]$$ $$h = x^m +b_1x^{m-1} + ... + b_m \in K[x]$$ such that$f = gh \in R[x]$but$h \notin R[x]$. Let$r:= \min {i | b_i \notin R}$. Since$f \in R[x]$we have$b_r + a_1b_{r-1} + ... + a_{r-1}b_1 + a_r \in R$, so$b_r \in R\$. It is a contradiction.