24. Let $p(x)=x^{2}+b x+c$, where $b$ and $c$ are integers. If $p(x)$ is a factor of both $x^{4}+6 x^{2}+25$ and $3 x^{4}+4 x^{2}+28 x+5$, then $p(1)$ is
(A) 0 ; (B) 1 ; (C) 2 ; (D) 4 ; (E) 8 .
$\frac{24}{\mathrm{D}}$  
24. Since $p(x)$ is a factor of $x^{4}+6 x^{2}+25$ and $3 x^{4}+4 x^{2}+28 x+5$, it is also a factor of $3\left(x^{4}+6 x^{2}+25\right)-\left(3 x^{4}+4 x^{2}+28 x+5\right)$. The latter equals $14 x^{2}-28 x+70$, i.e., $14\left(x^{2}-2 x+5\right)$. Therefore, $p(x)=x^{2}-2 x+5$, $p(1)=4$.