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Difference between revisions of "2023 AMC 10B Problems/Problem 12"
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==Solution==
 
==Solution==
  
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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.
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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 +,-,+,-,....,+.  There are 11 intervals, so there are 6 positives and 5 negatives.
  
 
~Technodoggo
 
~Technodoggo

Revision as of 16:51, 15 November 2023

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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