$2 \cdot 114$ Given the polynomial $a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}$, where $n$ is a positive integer or zero, $a_{0}$ is a positive integer, and the remaining $a$ are integers or zero, let $h=n+a_{0}+$ $\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right|$, when $h=3$, the number of polynomials that can be formed is
(A) 3 .
(B) 5 .
(C) 6 .
(D) 7 .
(E) 9 .
(10th American High School Mathematics Examination, 1959)
[Solution] Given $3=n+a_{0}+\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right|$, the above indeterminate equation has solutions:
$$
\begin{array}{l}
n=0, \quad a_{n}=3 ; \\
n=1, \quad a_{n-1}=2 \text {; } \\
n=1, \quad a_{n}= \pm 1, \quad a_{n-1}=1 \text {; } \\
n=2, \quad a_{n-2}=1, \quad a_{n-1}=a_{n}=0 . \\
\end{array}
$$

The corresponding polynomials are: $3 x^{0}, 2 x^{1}, x^{1}+1, x^{1}-1, x^{2}$. Therefore, the answer is $(B)$.