# 1959 AHSME Problems/Problem 48

Given the polynomial $a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$, where $n$ is a positive integer or zero, and $a_0$ is a positive integer. The remaining $a$'s are integers or zero. Set $h=n+a_0+|a_1|+|a_2|+\cdots+|a_n|$. [See example 25 for the meaning of $|x|$.] The number of polynomials with $h=3$ is: $\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 9$

## Solution

We perform casework by the value of $n$, the degree of our polynomial $a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$.

Case $n = 0$: In this case we are forced to set $a_0 = 3$. This contributes $1$ possibility.

Case $n = 1$: In this case we must have $a_0 + |a_1| = 2$, so our polynomial could be $1 + 1x, 0 + 2x, -1 + 1x$. This contributes $3$ possibilities.

Case $n = 2$: In this case we must have $a_0 + |a_1| + |a_2| = 1$. However, because $a_0$ must be positive, it has to be $1$, so our polynomial can only be $0 + 0x + 1x^2$. This contributes $1$ possibility.

Case $n\geq 3$: This case is impossible because $h = n+a_0+|a_1|+|a_2|+\cdots+|a_n|\geq n + a_0\geq 3 + 1 = 4 > 3$, so it contributes $0$ possibilities.

Adding the results from all four cases, we find that there are $1 + 3 + 1 + 0 = 5$ possibilities in total, so our answer is $\boxed{(B)}$.

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