Question

When the roots of the polynomial $P(x) = (x-1)^1(x-2)^2(x-3)^3 \cdots (x-10)^{10}$ are removed from the real number line, what remains is the union of $11$ disjoint open intervals. On how many of those intervals is $P(x)$ positive?

$3$

$7$

$6$

$4$

$5$

Answer: C

Concepts:polynomial parity inequality

Difficulty rating: 1560

Small Hint:

For $x \gt 10$ every factor is positive; the sign of $P$ can only change at a root

Big Hint:

Crossing the root $x = i$ flips the sign exactly when the exponent $i$ is odd

Solution:

For $x \gt 10,$ every factor $(x - i)^i$ is positive, so $P(x) \gt 0.$ Now move left. Crossing $x = i$ flips the sign only when $i$ is odd, that is at $i = 9, 7, 5, 3, 1.$ So the eleven intervals, right to left, carry signs $+, +, -, -, +, +, -, -, +, +, -.$ Six are positive. Therefore, the answer is C.

2023 AMC 10B Problem 12

Answer: C

Difficulty rating: 1560

Small Hint:

Big Hint:

Solution:

Problem 12 in Other Years