Let $f$ be a polynomial function such that, for all real $x$,
\[f(x^2 + 1) = x^4 + 5x^2 + 3.\]
For all real $x$, $f(x^2-1)$ is

$ \textbf{(A)}\ x^4+5x^2+1\qquad\textbf{(B)}\ x^4+x^2-3\qquad\textbf{(C)}\ x^4-5x^2+1\qquad\textbf{(D)}\ x^4+x^2+3\qquad\textbf{(E)}\ \text{None of these} $
1. We start with the given functional equation for the polynomial function \( f \):
   \[
   f(x^2 + 1) = x^4 + 5x^2 + 3
   \]
   We need to determine the form of \( f(x) \).

2. Let \( y = x^2 + 1 \). Then, we have:
   \[
   f(y) = (y-1)^2 + 5(y-1) + 3
   \]
   Expanding the right-hand side:
   \[
   (y-1)^2 = y^2 - 2y + 1
   \]
   \[
   5(y-1) = 5y - 5
   \]
   Adding these together:
   \[
   f(y) = y^2 - 2y + 1 + 5y - 5 + 3 = y^2 + 3y - 1
   \]
   Therefore, we have:
   \[
   f(x) = x^2 + 3x - 1
   \]

3. Now, we need to find \( f(x^2 - 1) \). Substitute \( x^2 - 1 \) into the polynomial \( f \):
   \[
   f(x^2 - 1) = (x^2 - 1)^2 + 3(x^2 - 1) - 1
   \]
   Expanding the terms:
   \[
   (x^2 - 1)^2 = x^4 - 2x^2 + 1
   \]
   \[
   3(x^2 - 1) = 3x^2 - 3
   \]
   Adding these together:
   \[
   f(x^2 - 1) = x^4 - 2x^2 + 1 + 3x^2 - 3 - 1
   \]
   Simplifying:
   \[
   f(x^2 - 1) = x^4 + x^2 - 3
   \]

4. Therefore, the expression for \( f(x^2 - 1) \) is:
   \[
   \boxed{x^4 + x^2 - 3}
   \]