AMC 10 · 2023 · #12
Grade 8 algebraProblem
When the roots of the polynomial
P(x)=(x−1)1(x−2)2(x−3)3⋅⋅⋅⋅(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?
Pick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The polynomial $P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdots (x-10)^{10}$ has roots at the integers $1, 2, \ldots, 10$, 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.
Givens: $P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdots (x-10)^{10}$; The exponent on $(x-k)$ is exactly $k$, for $k = 1, 2, \ldots, 10$; $11$ open intervals: $(-\infty, 1), (1, 2), \ldots, (9, 10), (10, \infty)$; Answer choices: (A) $3$, (B) $7$, (C) $6$, (D) $4$, (E) $5$
Unknowns: Number of intervals on which $P(x) > 0$
Understand
Restated: The polynomial $P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdots (x-10)^{10}$ has roots at the integers $1, 2, \ldots, 10$, 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.
Givens: $P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdots (x-10)^{10}$; The exponent on $(x-k)$ is exactly $k$, for $k = 1, 2, \ldots, 10$; $11$ open intervals: $(-\infty, 1), (1, 2), \ldots, (9, 10), (10, \infty)$; Answer choices: (A) $3$, (B) $7$, (C) $6$, (D) $4$, (E) $5$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #5 Look for a Pattern, #2 Make a Systematic List, #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
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 $P_3(x) = (x-1)(x-2)^2(x-3)^3$ first to verify the rule before applying to $10$ roots.
Execute — Answer: C
6.NS.C.6 Step 1 - Draw a number line.
- Mark the $10$ roots $1, 2, 3, \ldots, 10$ as dots.
- This carves out $11$ open intervals.
- The job is to label each interval with the sign of $P(x)$ on it.
💡 Putting roots on a number line turns an abstract sign question into a labeling exercise — Grade 6 number-line literacy.
8.EE.A.1 Step 2 - Notice what each factor $(x-k)^k$ contributes to the sign.
- If $k$ (the exponent) is even, $(x-k)^k$ is a power of a real number raised to an even integer, so it is always $\ge 0$ — and strictly $> 0$ inside any open interval avoiding $x = k$.
- Even-exponent factors NEVER contribute a negative sign.
- If $k$ is odd, $(x-k)^k$ has the same sign as $(x-k)$: positive when $x > k$, negative when $x < k$.
💡 Even powers are always nonnegative — Grade 8 exponent rule — so only odd-power factors can flip the sign of the whole product.
8.EE.A.1 Step 3 - List the roots whose exponent is odd (these are the only sign-flippers) and the ones whose exponent is even (these are sign-frozen).
- Odd exponents $k$: $1, 3, 5, 7, 9$.
- Even exponents $k$: $2, 4, 6, 8, 10$.
- So sign flips happen as $x$ crosses $1, 3, 5, 7, 9$; sign stays the same as $x$ crosses $2, 4, 6, 8, 10$.
💡 Separating roots by parity-of-exponent gives a clean tracking list — Tool #2's organizing power.
7.NS.A.2 Step 4 - Anchor the sign on the rightmost interval, then walk left.
- For $x > 10$, every factor $(x - k)$ is positive, so every $(x-k)^k > 0$, so $P(x) > 0$.
- Now sweep left, flipping only at odd-exponent roots.
💡 Pick a corner where the sign is obvious (everything positive), then carry sign across the line — Grade 7 sign-of-product reasoning.
8.EE.A.1 Step 5 - Walk right to left, applying the flip/no-flip rule from Step 3.
- Above the line, record "+" or "-" on each interval.
💡 Each transition is one step: flip the sign or keep it. Systematic listing avoids missing intervals.
8.EE.A.1 Step 6 - Count the intervals labeled "+": $(10, \infty), (9, 10), (6, 7), (5, 6), (2, 3), (1, 2)$.
- That is $6$ intervals.
💡 The matching choice (C) $= 6$ is the multiple-choice closeout.
6.NS.C.6 Draw a number line. Mark the $10$ roots $1, 2, 3, \ldots, 10$ as dots. This carv 8.EE.A.1 Notice what each factor $(x-k)^k$ contributes to the sign. If $k$ (the exponent) 8.EE.A.1 List the roots whose exponent is odd (these are the only sign-flippers) and the 7.NS.A.2 Anchor the sign on the rightmost interval, then walk left. For $x > 10$, every f 8.EE.A.1 Walk right to left, applying the flip/no-flip rule from Step 3. Above the line, 8.EE.A.1 Count the intervals labeled "+": $(10, \infty), (9, 10), (6, 7), (5, 6), (2, 3), Review
Reasonableness: 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 $(10, \infty)$ 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 $(-\infty, 1)$: $- \to + \to + \to - \to - \to + \to + \to - \to - \to + \to +$. That's $-,+,+,-,-,+,+,-,-,+,+$ — six "+" signs. Same answer. Quick sample-point check: $x = 1.5$ gives $(0.5)^1 (-0.5)^2 (-1.5)^3 \cdots$, 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.
Alternative: Tool #9 (Easier Problem): try $P_2(x) = (x-1)(x-2)^2$. Roots $1, 2$. Intervals: $(-\infty, 1), (1, 2), (2, \infty)$. Sign sweep from right: $(2, \infty) \to +$; cross $2$ (even, no flip) $\to (1, 2) +$; cross $1$ (odd, flip) $\to (-\infty, 1) -$. So $2$ positive intervals. Now $P_3(x) = (x-1)(x-2)^2(x-3)^3$: roots $1, 2, 3$. Sweep: $(3, \infty) +$; cross $3$ (odd, flip) $\to (2, 3) -$; cross $2$ (even) $\to (1, 2) -$; cross $1$ (odd, flip) $\to (-\infty, 1) +$. So $2$ positive intervals. Pattern check: the answer for $n$ roots is the count of "+" signs in the alternation governed by odd-exponent roots only. For $n = 10$ with $5$ odd-exponent roots, that count is $6$. Matches.
CCSS standards used (min grade 8)
6.NS.C.6Understand a rational number as a point on the number line (Setting up the $10$-root number line that produces the $11$ open intervals to be classified.)8.EE.A.1Know and apply the properties of integer exponents (Using "even exponent → nonnegative power, odd exponent → preserves base sign" to reduce sign tracking to only odd-multiplicity roots.)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Anchoring the sign on $(10, \infty)$ as positive (product of positives) and propagating the sign across each root.)
⭐ 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.
⭐ 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.