AMC 10 · 2023 · #12

Grade 8 algebra

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

📏 Medium solution 💡 3 insights

Problem

The polynomial P(x) = (x-1)¹ (x-2)² (x-3)³ … (x-10)¹⁰ has roots at the integers 1, 2, …, 10, each with the multiplicity shown by its exponent. Remove those 10 points from the number line; what remains is 11 open intervals. Count how many of those intervals make P(x) positive.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Draw a number line and mark roots 1–10 as dots; they carve the line into 11 open intervals, each to be labeled with the sign of P(x).

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

2STEP 2

An even exponent keeps (x−k)^k from ever going negative, so only the odd-exponent factors can flip the sign of the whole product.

(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

Split roots by exponent parity: the sign flips only at the odd roots 1, 3, 5, 7, 9 and stays put at the even roots 2, 4, 6, 8, 10.

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

4STEP 4

Anchor the far-right interval: for x ＞ 10 every factor is positive, so P(x) ＞ 0 on (10, ∞); now sweep left.

Anchor: P(x) ＞ 0 on (10, ∞)

5STEP 5

Walk right to left, applying the flip-at-odd rule, and record + or − above each interval.

(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 plus intervals — (1, 2), (2, 3), (5, 6), (6, 7), (9, 10), (10, ∞) — giving 6 positive intervals, choice (C).

#{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 10 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.