AMC 10 · 2022 · #21

Grade 8 number-theory

polynomial-factoringpolynomial-rootssystems-of-equationspolynomial-remainderconvert-to-algebra easier-related-problemconvert-to-algebraidentify-subproblems ↑ Prerequisites: polynomial-factoringsystems-of-equations

📏 Long solution 💡 4 insights

Problem

Find a polynomial P(x) with rational coefficients that leaves remainder x+2 when divided by x²+x+1 and remainder 2x+1 when divided by x²+1. Among all such polynomials, pick the unique one with the smallest possible degree, then compute the sum of the squares of its coefficients.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

Tool #9 (Easier Problem) — first try the smallest possible degree for P (degree 2, with constant quotients) to see what goes wrong; then bump up to degree 3 (linear quotients). Tool #7 (Subproblems) — splitting equation A = B into three coefficient-equations (one per power of x) turns the polynomial puzzle into a small system. Tool #13 (Algebra) — solve that tiny system for the unknown coefficients. Tool #3 (Eliminate) — quickly verify the final sum 1²+2²+3²+3² = 23 matches choice (E).

1STEP 1

If P has degree 2, both quotients are constants; matching terms forces 3 = 2, a contradiction, so degree 2 is impossible.

degree 2 → 3 = 2 (impossible)

2STEP 2

For degree 3 the quotients are linear; matching the leading term forces the same a, so Q₁ = ax+b and Q₂ = ax+c.

Q₁ = ax+b, Q₂ = ax+c

3STEP 3

Expand form 1: (x²+x+1)(ax+b) + (x+2) = ax³ + (a+b)x² + (a+b+1)x + (b+2).

P = ax³ + (a+b)x² + (a+b+1)x + (b+2)

4STEP 4

Expand form 2: (x²+1)(ax+c) + (2x+1) = ax³ + cx² + (a+2)x + (c+1).

P = ax³ + cx² + (a+2)x + (c+1)

5STEP 5

Matching coefficients term by term (x, constant, then x²) solves the system to a = 1, b = 1, c = 2.

a = 1, b = 1, c = 2

6STEP 6

Substitute a=1, b=1 back into form 1: P(x) = x³ + 2x² + 3x + 3, and form 2 confirms the same polynomial.

P(x) = x³ + 2x² + 3x + 3

7STEP 7

Square the coefficients 1, 2, 3, 3 and add: 1² + 2² + 3² + 3² = 1 + 4 + 9 + 9 = 23 → (E).

1² + 2² + 3² + 3² = 23 → (E)

Answer

Sanity. Verify P(x) = x³ + 2x² + 3x + 3 in both original conditions. Divide by x² + x + 1: P = (x²+x+1)(x+1) + (x+2), remainder x+2 — matches. Divide by x² + 1: P = (x²+1)(x+2) + (2x+1), remainder 2x+1 — matches. Sum of squared coefficients 1+4+9+9 = 23, which is exactly choice (E). The degree-2 attempt failed by a numerical contradiction (3 = 2), so degree 3 is genuinely the least. The choices 10, 13, 19, 20, 23 are decoys for forgetting a coefficient or for the unsquared sum 1+2+3+3 = 9.

💡Key takeaway

This AMC 10 problem only needs Grade 8 systems of equations you already know — try the smallest possible degree first (it fails by a clean 3=2 contradiction), then write the next degree using letters for the unknown coefficients and match like terms to set up three tiny equations. Solve to get P(x) = x³ + 2x² + 3x + 3, and the sum of squares of coefficients is 1+4+9+9 = 23.