AMC 10 · 2023 · #6

Grade 8 algebra

Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type

polynomial-rootssign-analysisparitypattern-recognition sign-analysispattern-recognitioncasework ↑ Prerequisites: polynomial-rootsparity

📏 Medium solution 💡 3 insights

Problem

A polynomial has the integers one through ten as roots, each with multiplicity equal to the number itself. Removing those ten points from the number line leaves eleven open intervals. Count how many of them make the polynomial positive.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram): the sign of P(x) lives naturally on a number line with 10 marked roots — draw it, label "+" or "-" above each interval. Tool #5 (Pattern) catches the key shortcut: (x-k)^k flips the overall sign at x = k only when the exponent k is odd; even powers can never go negative. So only odd roots matter for sign flips. Tool #2 (Systematic List) walks the 11 intervals from right (where P(x) ＞ 0 obviously) to left, marking each. Tool #9 (Easier Problem) sanity-check: try P₃(x) = (x-1)(x-2)²(x-3)³ first to verify the rule before applying to 10 roots.

1STEP 1

List the intervals

List the eleven intervals.

Intervals: (-∞,1),(1,2),(2,3),…,(9,10),(10,∞)

2STEP 2

Find where the sign flips

At even multiplicity it does not flip.

(x-k)^k ＞ 0 for all x ≠ k when k even; (x-k)^k has the sign of (x-k) when k odd.

3STEP 3

List the flip points

Only the odd roots flip.

Flip at: 1, 3, 5, 7, 9; No flip at: 2, 4, 6, 8, 10

4STEP 4

Anchor at the far right

The far right is positive.

Anchor: P(x) ＞ 0 on (10, ∞)

5STEP 5

Sweep leftward

Fill in the signs moving left.

(10,∞){:}+ | 10:no flip | (9,10){:}+ | 9:flip | (8,9){:}- | 8:no flip | (7,8){:}- | 7:flip | (6,7){:}+ | 6:no flip | (5,6){:}+ | 5:flip | (4,5){:}- | 4:no flip | (3,4){:}- | 3:flip | (2,3){:}+ | 2:no flip | (1,2){:}+ | 1:flip | (-∞,1){:}-

6STEP 6

Count the positive intervals

There are 6 positive intervals.

#{positive intervals} = 6 → (C)

Answer

Sanity-check via parity: of the 11 intervals, 5 are bounded by two odd-exponent roots, but the simpler check is symmetric splitting. Starting from + on (10, ∞) and flipping at 5 roots (1, 3, 5, 7, 9), the sign alternates in blocks framed by those flips. Walking left to right starting from "-" on (-∞, 1): - → + → + → - → - → + → + → - → - → + → +. That's -,+,+,-,-,+,+,-,-,+,+ — six "+" signs. Same answer. Quick sample-point check: x = 1.5 gives (0.5)¹ (-0.5)² (-1.5)³ …, where (0.5) and (-1.5)³ = -3.375, (-2.5)⁵ negative, (-3.5)⁷ negative, (-4.5)⁹ negative; even powers all positive. Product of one positive (0.5) and four negatives (1, 3, 5, 7, 9 above 1.5, but 1 doesn't apply since x ＞ 1; for x = 1.5 the odd-exponent roots with x ＜ k are 3, 5, 7, 9 — four negatives → positive product). Confirms P(1.5) ＞ 0, matching (1, 2) being positive.

💡Key takeaway

This AMC 12 problem only needs Grade 8 "even powers stay nonnegative" you already know — that rule says only the odd-exponent roots (1, 3, 5, 7, 9) flip the sign of P(x), and walking the number line from right to left gives exactly 6 positive intervals.