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| − | When the roots of the polynomial
| + | #redirect[[2023 AMC 12B Problems/Problem 6]] |
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| − | <math>P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}</math>
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| − | are removed from the number line, what remains is the union of 11 disjoint open intervals. On how many of these intervals is <math>P(x)</math> positive?
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| − | ==Solution==
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| − | The interval of the alternating signs. <math>P(x)</math> is a product of <math>(x-r_n)</math> or 10 terms. When <math>x < 1</math>, all terms are <math>< 0</math>, but <math>P(x) > 0</math> because they are even number of terms. The sign keep alternates <math>+,-,+,-,....,+</math>. There are 11 intervals, so there are 6 positives and 5 negatives.
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| − | ~Technodoggo
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