Consider two polynomials $A(X), B(X)$ both $\in \mathbb{Z}[X]$ such that $\deg (B) < \deg(A)$. And let us perform a long polynomial division of $A(X)$ over $B(X)$, $$A(X)=Q(X)B(X)+R(X)$$ $$Q(X)=\left\lfloor\frac{A(X)}{B(X)}\right\rfloor \quad \mbox{is the quotient} \quad \mbox{and $R(x)$ is the remainder}$$ Now consider the variable $X$ taking positive integer value , say $X=n$ and perform the Euclidean division of $A(n)$ over $B(n)$, we get $A(n)=q(n)B(n)+r(n)$ where $q(n)$ and $r(n)$ are the usual quotient and remainder. What we can get is there is not always agreement between Long polynomial division and Euclidean divison when the variable takes numerical integer values. Namely $$Q(n) \not= q(n) \quad \mbox{and} \quad R(n)\not= r(n)$$ To fix ideas here is an example: Fix a positive integer $N$ and consider $$A(X)=X^2-1, \quad B(X)=X+N$$ so that synthetic or polynomial division gives: $$Q(X)=X-N \quad \mbox{and} \quad R(X)=N^2-1$$ It is clear for this example that both divisions do not agree for $n<N^2-N-1$ we even have a negative numerical quotient $Q(n)$ for $1<n<N$ , while both $A(n)$ and $B(n)$ are positive integers.
I am interested in finding a general way (if it exist) which yield an algebraic expression of new quotient and remainder $Q^{'}(X)$ and $R^{'}(X)$ such that $A(X)=Q^{'}(X)B(X)+R^{'}(X)$ and $Q^{'}(n)=q(n), R^{'}(n)=r(n)$, restricted to the simplest case where $\deg(A)=2$, $\deg(B)=1$.