Let $p(x) = x^{2} + bx + c$, where $b$ and $c$ are integers. 
If $p(x)$ is a factor of both $x^{4} + 6x^{2} + 25$ and $3x^{4} + 4x^{2} + 28x + 5$, what is $p(1)$?
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ 8$
$p(x)$ must be a factor of $3(x^4+6x^2+25)-(3x^4+4x^2+28x+5)=14x^2-28x+70=14(x^2-2x+5)$.
Therefore $p(x)=x^2 -2x+5$ and $p(1)=4$.
The answer is $\fbox{(D) 4}$.