2023 AMC 10B Problems/Problem 12

When the roots of the polynomial

are removed from the number line, what remains is the union of 11 disjoint open intervals. On how many of these intervals is positive?

Solution

The interval of the alternating signs. is a product of or 10 terms. When , all terms are , but because they are even number of terms. The sign keep alternates +,-,+,-,....,+. There are 11 intervals, so there are 6 positives and 5 negatives.

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