2023 AMC 10B Problems/Problem 12

When the roots of the polynomial

$P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}$

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

Solution

The interval of the alternating signs. $P(x)$ is a product of $(x-r_n) or 10 terms$ . When $x < 1$ , all terms are $< 0$ , but $P(x) > 0$ because they are even number of terms. The sign keep alternates +,-,+,-,....,+. There are 11 intervals, so there are 6 positives and 5 negatives.

~Technodoggo

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2023 AMC 10B Problems/Problem 12

Solution