AMC 10 · 2022 · #7

Grade 7 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

polynomial-factoringpolynomial-rootsvieta-formulasfactorssystematic-enumeration systematic-enumerationpattern-recognitionidentify-subproblems ↑ Prerequisites: polynomial-factoringfactors

📏 Medium solution 💡 2 insights

Problem

For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 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.

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Toolkit + CCSS Solution

Understand

Restated: For how many integer values of the constant $k$ does the quadratic $x^2 + kx + 36$ have two distinct integer roots?

Givens: The polynomial is $x^2 + kx + 36$; The two roots must be integers and distinct (not equal to each other); Answer choices: (A) $6$, (B) $8$, (C) $9$, (D) $14$, (E) $16$

Unknowns: The number of integer values of $k$ that make the two roots distinct integers

Understand

Restated: For how many integer values of the constant $k$ does the quadratic $x^2 + kx + 36$ have two distinct integer roots?

Givens: The polynomial is $x^2 + kx + 36$; The two roots must be integers and distinct (not equal to each other); Answer choices: (A) $6$, (B) $8$, (C) $9$, (D) $14$, (E) $16$

Plan

Primary tool: #2 Make a Systematic List

Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems

The roots have to multiply to $36$, so we are really being asked to list every integer factor pair of $36$ in which the two factors are different. Tool #2 (Systematic List) is built for 'find every pair' with no duplicates. Tool #7 (Subproblems) splits the work into 'list positive factor pairs', 'list negative factor pairs', and 'count distinct $k$'. Tool #9 (Easier Problem) is the bridge: instead of solving for $k$ first, solve the simpler factor-pair question and read $k$ off at the end. We avoid the quadratic formula — factor-pair counting is the natural elementary path.

Execute — Answer: B

#7 Identify Subproblems 7.EE.A.1 Step 1

Match the polynomial with its factored form.
If the two integer roots are $r$ and $s$, then $x^2 + kx + 36 = (x - r)(x - s)$.
Expanding the right side: $(x - r)(x - s) = x^2 - (r + s)x + rs$.
Comparing coefficients: $rs = 36$ and $r + s = -k$.

$$x^2 + kx + 36 = (x-r)(x-s) \;\Rightarrow\; rs = 36,\;r + s = -k$$

💡 Expanding $(x-r)(x-s)$ shows the constant term is $rs$ and the middle coefficient is the negative sum — this turns the problem into a factor-pair hunt.

#2 Make a Systematic List 4.OA.B.4 Step 2

List every positive factor pair $(r, s)$ of $36$ with $r < s$ (writing $r < s$ avoids listing $(a, b)$ and $(b, a)$ as two cases since they give the same $k$).
The positive divisors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18, 36$.
Pair them off: $(1, 36), (2, 18), (3, 12), (4, 9)$.
The pair $(6, 6)$ would have equal roots, so drop it.

$$\text{positive pairs with}\;r<s:\;(1, 36), (2, 18), (3, 12), (4, 9)$$

💡 Listing factor pairs in size order guarantees no repeats and no misses.

#2 Make a Systematic List 6.NS.C.6 Step 3

Negative factor pairs come from flipping both signs: if $r \cdot s = 36$, the pair $(-r, -s)$ also multiplies to $36$ because $(-)\cdot(-) = (+)$.
So mirror the four positive pairs: $(-1, -36), (-2, -18), (-3, -12), (-4, -9)$.
The mirror of $(6, 6)$ would be $(-6, -6)$ — also dropped (still equal roots).

$$\text{negative pairs}:\;(-1, -36), (-2, -18), (-3, -12), (-4, -9)$$

💡 Two negatives multiply to a positive — the negative side of the rational number line gives a second clean batch of pairs.

#7 Identify Subproblems 6.NS.C.6 Step 4

Mixed-sign pairs (one positive, one negative) are impossible: a positive times a negative is negative, but we need the product to be $+36$.
So no other pairs exist.

$$(+) \times (-) = (-) \ne 36$$

💡 Sign rules block any 'mixed' pair from ever multiplying to a positive $36$.

#2 Make a Systematic List 7.NS.A.1 Step 5

Convert each pair to its $k = -(r+s)$.
Positive pairs give negative $k$: $(1,36)\to k=-37$; $(2,18)\to k=-20$; $(3,12)\to k=-15$; $(4,9)\to k=-13$.
Negative pairs give positive $k$: $(-1,-36)\to k=37$; $(-2,-18)\to k=20$; $(-3,-12)\to k=15$; $(-4,-9)\to k=13$.

$$k \in \{-37, -20, -15, -13,\;37, 20, 15, 13\}$$

💡 Each factor pair fixes $r + s$, and $k$ is just the negation of that sum.

#2 Make a Systematic List 4.NBT.A.2 Step 6

All eight values of $k$ are distinct (the negatives and positives are mirror images of each other, none of them coincide).
Count: $\mathbf{8}$ values.
Answer (B).

$$|\{-37, -20, -15, -13, 13, 15, 20, 37\}| = 8 \;\Rightarrow\; \textbf{(B)}$$

💡 Comparing the eight integers shows no duplicates — straight count of a small finite list.

[1] #7 7.EE.A.1 Match the polynomial with its factored form. If the two integer roots are $r$ an

[2] #2 4.OA.B.4 List every positive factor pair $(r, s)$ of $36$ with $r < s$ (writing $r < s$ a

[3] #2 6.NS.C.6 Negative factor pairs come from flipping both signs: if $r \cdot s = 36$, the pa

[4] #7 6.NS.C.6 Mixed-sign pairs (one positive, one negative) are impossible: a positive times a

[5] #2 7.NS.A.1 Convert each pair to its $k = -(r+s)$. Positive pairs give negative $k$: $(1,36)

[6] #2 4.NBT.A.2 All eight values of $k$ are distinct (the negatives and positives are mirror ima

Review

Reasonableness: Spot-verify the largest $k$: if $k = -37$ the polynomial is $x^2 - 37x + 36 = (x - 1)(x - 36)$ with roots $1$ and $36$ ✓. Spot-verify a positive $k$: if $k = 13$ then $x^2 + 13x + 36 = (x + 4)(x + 9)$ with roots $-4$ and $-9$ ✓. The structural reason the count is $8$ and not $9$: the factor pair $(6, 6)$ — which would give $k = -12$ — is forbidden because the roots must be distinct. The structural reason it is not $14$ or $16$: pairs like $(1, 36)$ and $(36, 1)$ are the same pair (and give the same $k$), so unordered pair counting is correct.

Alternative: Tool #3 (Eliminate by plugging in answer choices): The number of positive divisors of $36$ is $9$ ($1, 2, 3, 4, 6, 9, 12, 18, 36$). That gives at most $9$ positive factor pairs counted as unordered, but we drop the equal pair $(6, 6)$, leaving $4$. Double for negative signs: $4 \times 2 = 8$. Choice (B) is the only one consistent with 'must drop the square root pair'. This shortcut skips writing out every $k$, but the systematic list above also verifies they really are eight distinct integers.

CCSS standards used (min grade 7)

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions (Expanding $(x - r)(x - s) = x^2 - (r+s)x + rs$ to read off $rs = 36$ and $r + s = -k$.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Listing the four positive factor pairs of $36$ with $r < s$.)
6.NS.C.6 Understand a rational number as a point on the number line (Extending factor pairs into the negative integers using sign rules.)
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers (Computing $k = -(r+s)$ for each pair, including negative sums.)
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Confirming the eight resulting $k$-values are all distinct.)

⭐ This AMC 10 problem only needs Grade 7 expanding $(x-r)(x-s)$ you already know — the constant $36$ has to come from $r \cdot s$, so list the four positive factor pairs of $36$ with $r \ne s$, double them for negatives, and read off $8$ values of $k$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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