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2023 AMC 10B Problem 12

2023 AMC 10B Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Polynomials · difficulty 3 of 5 · about 3 min · position P11-15

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?

A)
$3$
B)
$7$
C)
$6$
D)
$4$
E)
$5$

Answer

Key insight

P is positive for x > 10 and its sign flips only when crossing a root of odd multiplicity; track the flips at 9, 7, 5, 3, 1.

Solution

For $x > 10$ every factor $x - k$ is positive, so $P(x) > 0$ on $(10, \infty)$ .

Move left across the roots one at a time. Crossing $x = k$ changes the sign of the single factor $(x - k)$ , which appears $k$ times, so the sign of $P$ flips exactly when $k$ is odd and stays the same when $k$ is even.

Starting from $+$ on $(10,\infty)$ :

interval	$(9,10)$	$(8,9)$	$(7,8)$	$(6,7)$	$(5,6)$	$(4,5)$	$(3,4)$	$(2,3)$	$(1,2)$	$(-\infty,1)$
root crossed	10	9	8	7	6	5	4	3	2	1
sign	$+$	$-$	$-$	$+$	$+$	$-$	$-$	$+$	$+$	$-$

The positive intervals are $(10,\infty)$ , $(9,10)$ , $(6,7)$ , $(5,6)$ , $(2,3)$ , $(1,2)$ : six in all.

The answer is $\boxed{\textbf{(C)}\ 6}$ .

Why this works

A polynomial's sign between consecutive roots is constant, and the only thing that can change it is passing a root, where the sign flips if and only if the multiplicity is odd. Anchoring at the far right (where every factor is positive) and walking left with the parity rule produces the whole sign chart mechanically.

Alternative approach

Pair the roots: crossing an even root does nothing and crossing the odd root just below it flips the sign, so the sign is constant on each block $(2j-1, 2j) \cup (2j, 2j+1)$ and alternates from block to block. From the right the blocks are $+, -, +, -, +$ (with $(-\infty, 1)$ negative at the end): three positive blocks of two intervals each, so $6$ .

The trap

Starting the sign chart from the far left assuming P is positive there; the degree 55 is odd, so P is negative for x < 1.

Common mistakes

Starting the sign chart from the far left assuming P is positive there; the degree 55 is odd, so P is negative for x < 1.
Flipping the sign at every root, ignoring multiplicities (this happens to give $6$ here, but for the wrong reason and fails in general).

Techniques

Use an invariant, parity, or coloring argument · Compute small cases, spot the pattern, generalize