Polynomial divisibility is unchanged by coeff. field extension
Solution 1:
Hint $ $ It follows from the uniqueness of the quotient (and remainder) in the division algorithm (which is the same in $\,K[x]\,$ and $\,E[x],\,$ using the polynomial degree as the Euclidean valuation). Thus since dividing $\,f\,$ by $\,g\,$ in $\,E[x]\,$ leaves remainder $0$, by uniqueness, the remainder must also be $\,0\,$ in $\,K[x]\,,\,$ i.e. $\,g\ |\ f\ $ in $\,E[x]\,$ $\Rightarrow\, g\ |\ f\ $ in $\,K[x].\,$
This is but one of many examples of the power of uniqueness theorems for proving equalities.