2025 AMC 12A Problem 25

This problem was voided: as written, its answer is not among the choices. Here is the count. For $\{f \le 0\} = [a, b] \cup (c, d),$ the endpoints $a, b$ of the closed interval must be zeros of $P$ at which $Q\ne0,$ while $c,d$ must be poles. The required sign pattern is therefore that of $$g(x)=\frac{(x-a)(x-b)}{(x-c)(x-d)}.$$ The unused third factor of $P$ must match the unused third factor of $Q;$ otherwise there would be an extra zero, pole, or sign change. If the common factor is the fifth value not among $a,b,c,d,$ it cannot make a hole inside either interval. In the ordered list of five values it may therefore occur before $a,$ between $b$ and $c,$ or after $d,$ giving $3$ functions. Alternatively, the common factor can equal $c$ or $d.$ For each of the $\binom54=5$ choices of $a,b,c,d,$ these two polynomial pairs simplify to the same formula $g$ and have the same domain (both omit $c,d$), so they define only one function. This gives $5$ more functions, for a literal total of $3+5=8.$ Counting the two polynomial pairs separately in each of the last five cases gives $3+2\cdot5=13,$ the provisional answer (E), but that does not answer the stated question about functions. Thus none of the printed choices is correct.