Let $F$ be a field, and $E$ be an extension of it. Let $p$ be a polynomial over $F$. We need to show that $$E \otimes_F \frac{F[x]}{(p)} \cong \frac{E[x]}{(p)}.$$
Here is what I thought : Consider the map $ E \times \frac{F[x]}{(p)} \to \frac{E[x]}{(p)} $ given by $(e, f + (p)) \mapsto ef + (p)$. This map is bilinear, thus by Universal property there exist an $F$-map $\phi:E \otimes_F \frac{F[x]}{(p)} \to \frac{E[x]}{(p)}$ such that $\phi(e \otimes f+ (p)) = ef + (p).$ I think that the map $\phi$ is an isomorphism, but I am not able to think of what the inverse is.
I feel that the above map is a natural bilinear map from $ E \times \frac{F[x]}{(p)} \to \frac{E[x]}{(p)} $, but in the other direction, if I have $g + (p) \in \frac{E[x]}{(p)},$ from $g$, I need to find a polynomial $f \in F[x]$ and an element $e \in E$ such that $g +(p) \mapsto e \otimes f+(p)$ is well defined and an $F$-map. I am not able to figure out any such natural map. Any help is appreciated.