If $A\subset B$, $\mathfrak{p}$ be a prime of $A$, $x+\mathfrak{p}B$ is a unit, irreducible $f\in A[X]$ has $f(x)=0$, must $[X^0]f\notin\mathfrak{p}$?

Question

Let $A$ be a commutative ring with unity, $B$ be an integral extension of $A$ (that is, every element in $B$ is a root of a monic polynomial in $A[X]$). Let $\mathfrak{p}$ be a prime ideal of $A$. Suppose that $x\in B$, and that $x+\mathfrak{p}B$ is a unit in the ring $B/\mathfrak{p}B$ ($\mathfrak{p}B$ is the ideal of $B$ generated by $\mathfrak{p}$). If a polynomial $f(X) = X^n + a_{n-1}X^{n-1} + \cdots + a_1X + a_0\in A[X]$, irreducible over $A[X]$ has $f(x)=0$, must we have $a_0\notin \mathfrak{p}$?