$29 \cdot 48$ 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.
(37th American High School Mathematics Examination, 1986)
[Solution] $x^{4}+6 x^{2}+25=\left(x^{4}+10 x^{2}+25\right)-4 x^{2}$
$=\left(x^{2}+5\right)^{2}-(2 x)^{2}$
$=\left(x^{2}+2 x+5\right)\left(x^{2}-2 x+5\right)$,

Also, $3 x^{4}+4 x^{2}+28 x+5=\left(x^{2}-2 x+5\right)\left(3 x^{2}+6 x+1\right)$,
then
$p(x)=x^{2}-2 x+5$, and $p(1)=4$.
Therefore, the answer is $(D)$.