For $x^2+2x+5$ to be a factor of $x^4+px^2+q$, the values of $p$ and $q$ must be, respectively:

$ \textbf{(A)}\ -2, 5\qquad\textbf{(B)}\ 5, 25\qquad\textbf{(C)}\ 10, 20\qquad\textbf{(D)}\ 6, 25\qquad\textbf{(E)}\ 14, 25 $
To determine the values of \( p \) and \( q \) such that \( x^2 + 2x + 5 \) is a factor of \( x^4 + px^2 + q \), we start by expressing \( x^4 + px^2 + q \) as a product of \( x^2 + 2x + 5 \) and another quadratic polynomial \( Q(x) \).

Let \( Q(x) = ax^2 + bx + c \). Then we have:
\[
(x^2 + 2x + 5)(ax^2 + bx + c) = x^4 + px^2 + q
\]

Expanding the left-hand side:
\[
(x^2 + 2x + 5)(ax^2 + bx + c) = ax^4 + bx^3 + cx^2 + 2ax^3 + 2bx^2 + 2cx + 5ax^2 + 5bx + 5c
\]

Combining like terms:
\[
ax^4 + (b + 2a)x^3 + (c + 2b + 5a)x^2 + (2c + 5b)x + 5c
\]

Since the right-hand side must equal \( x^4 + px^2 + q \), we equate the coefficients of corresponding powers of \( x \):
\[
\begin{aligned}
&\text{Coefficient of } x^4: & a &= 1 \\
&\text{Coefficient of } x^3: & b + 2a &= 0 \\
&\text{Coefficient of } x^2: & c + 2b + 5a &= p \\
&\text{Coefficient of } x: & 2c + 5b &= 0 \\
&\text{Constant term: } & 5c &= q \\
\end{aligned}
\]

Now, we solve these equations step-by-step:

1. From \( a = 1 \):
\[
a = 1
\]

2. From \( b + 2a = 0 \):
\[
b + 2(1) = 0 \implies b = -2
\]

3. From \( 2c + 5b = 0 \):
\[
2c + 5(-2) = 0 \implies 2c - 10 = 0 \implies 2c = 10 \implies c = 5
\]

4. From \( c + 2b + 5a = p \):
\[
5 + 2(-2) + 5(1) = p \implies 5 - 4 + 5 = p \implies p = 6
\]

5. From \( 5c = q \):
\[
5(5) = q \implies q = 25
\]

Thus, the values of \( p \) and \( q \) are \( 6 \) and \( 25 \), respectively.

The final answer is \(\boxed{(6, 25)}\).