AMC 10 · 2023 · #12

Grade 8 algebra

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

📏 Medium solution 💡 3 insights

Problem

When the roots of the polynomial

$P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot \cdot (x-10)^{10}$

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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

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

#1 Draw a Diagram 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.

$$\text{Intervals: } (-\infty,1),(1,2),(2,3),\ldots,(9,10),(10,\infty)$$

💡 Putting roots on a number line turns an abstract sign question into a labeling exercise — Grade 6 number-line literacy.

#5 Look for a Pattern 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 $x < k$.

$(x-k)^{k} > 0$ for all $x \ne k$ when $k$ even; $(x-k)^k$ has the sign of $(x-k)$ when $k$ odd.

💡 Even powers are always nonnegative — Grade 8 exponent rule — so only odd-power factors can flip the sign of the whole product.

#2 Make a Systematic List 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 $2, 4, 6, 8, 10$.

$$\text{Flip at: } 1, 3, 5, 7, 9; \quad \text{No flip at: } 2, 4, 6, 8, 10$$

💡 Separating roots by parity-of-exponent gives a clean tracking list — Tool #2's organizing power.

#1 Draw a Diagram 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.

$$\text{Anchor: } P(x) > 0 \text{ on } (10, \infty)$$

💡 Pick a corner where the sign is obvious (everything positive), then carry sign across the line — Grade 7 sign-of-product reasoning.

#2 Make a Systematic List 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.

$$(10,\infty){:}+ \;|\; 10:\text{no flip} \;|\; (9,10){:}+ \;|\; 9:\text{flip} \;|\; (8,9){:}- \;|\; 8:\text{no flip} \;|\; (7,8){:}- \;|\; 7:\text{flip} \;|\; (6,7){:}+ \;|\; 6:\text{no flip} \;|\; (5,6){:}+ \;|\; 5:\text{flip} \;|\; (4,5){:}- \;|\; 4:\text{no flip} \;|\; (3,4){:}- \;|\; 3:\text{flip} \;|\; (2,3){:}+ \;|\; 2:\text{no flip} \;|\; (1,2){:}+ \;|\; 1:\text{flip} \;|\; (-\infty,1){:}-$$

💡 Each transition is one step: flip the sign or keep it. Systematic listing avoids missing intervals.

#3 Eliminate Possibilities 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.

$$\#\{\text{positive intervals}\} = 6 \;\Rightarrow\; \textbf{(C)}$$

💡 The matching choice (C) $= 6$ is the multiple-choice closeout.

[1] #1 6.NS.C.6 Draw a number line. Mark the $10$ roots $1, 2, 3, \ldots, 10$ as dots. This carv

[2] #5 8.EE.A.1 Notice what each factor $(x-k)^k$ contributes to the sign. If $k$ (the exponent)

[3] #2 8.EE.A.1 List the roots whose exponent is odd (these are the only sign-flippers) and the

[4] #1 7.NS.A.2 Anchor the sign on the rightmost interval, then walk left. For $x > 10$, every f

[5] #2 8.EE.A.1 Walk right to left, applying the flip/no-flip rule from Step 3. Above the line,

[6] #3 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.6 Understand 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.1 Know 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.2 Apply 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.