|
|
| (11 intermediate revisions by 3 users not shown) |
| Line 1: |
Line 1: |
| − | When the roots of the polynomial
| + | #redirect[[2023 AMC 12B Problems/Problem 6]] |
| − | | |
| − | <math>P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}</math>
| |
| − | | |
| − | 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?
| |
| − | | |
| − | ==Solution==
| |
| − | | |
| − | The interval of the alternating signs. <math>P(x)</math> is a product of <math>(x-r_n) or 10 terms</math>. 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 +,-,+,-,....,+. There are 11 intervals, so there are 6 positives and 5 negatives.
| |
| − | | |
| − | ~Technodoggo
| |
