AMC 10 · 2014 · #16

Grade 9 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation
systems-of-equationsfunction-evaluationsymmetry-argumentpolynomial-substitution symmetry-argumentconvert-to-algebra ↑ Prerequisites: systems-of-equationsfunction-evaluation
📏 Long solution 💡 4 insights
Problem
Three values are given for a cubic, one short of pinning it down. Find a combination that is nevertheless determined.

Pick an answer.

(A)
0
(B)
k
(C)
6k
(D)
7k
(E)
14k
How to solve
Strategy Organize Information in More Ways

Chasing all four coefficients is hopeless, because three equations cannot determine four numbers. The way out is to notice how the data is arranged. Every input comes in a plus-minus pair around 0, and the target is a plus-minus pair too. So instead of studying P(x), study P(x) + P(-x). In that combination the odd-degree terms cancel, leaving only the x² and constant terms. The unknown leading coefficient — the one the data can never reach — is precisely one of the terms that cancels. What is left has two unknowns, and the givens supply exactly the two facts needed to fix them.

1STEP 1

Name the four coefficients

Four coefficients describe the cubic.

P(x) = Ax³ + Bx² + Cx + D, D = k ; A + B + C + D = 2k ; -A + B - C + D = 3k
2STEP 2

Add P(x) and P(-x)

Adding the mirror kills the odd terms.

P(-x) = -Ax³ + Bx² - Cx + D ⟹ P(x) + P(-x) = 2Bx² + 2D
3STEP 3

Read 2D and 2B off the givens

The givens supply exactly the two survivors.

2D = 2P(0) = 2k, 2B + 2D = P(1) + P(-1) = 2k + 3k = 5k ⟹ 2B = 3k
4STEP 4

Evaluate the identity at x = 2

Evaluating the identity gives 14k.

P(2) + P(-2) = 2B(2)² + 2D = 4(2B) + 2D = 4(3k) + 2k = 14k
5STEP 5

Build an actual cubic and confirm

An actual cubic confirms it, choice (D).

P(x) = x³ + 3k/2x² - (k/2 + 1)x + k: P(0) = k, P(1) = 2k, P(-1) = 3k, P(2) + P(-2) = (6k + 6) + (8k - 6) = 14k
Answer
14k
Two independent checks agree. First, the explicit cubics: with k = 1 and A = 1, P(x) = x³ + 1.5x² - 1.5x + 1 gives P(0) = 1, P(1) = 2, P(-1) = 3 as required, and P(2) + P(-2) = 12 + 2 = 14 = 14k; with A = -2 the same k gives P(2) + P(-2) = -6 + 20 = 14 again. Second, the wrong answers are exactly the predictable slips, which is a good sign the reasoning found the real mechanism. Choice (D) 7k is 4B + D, the value of the even part itself — the answer for a student who computes the average instead of the sum. Choice (C) 6k is the value of P(2) for the one polynomial through the three points that has no x³ term at all; that polynomial is the quadratic 3k/2x² - k/2x + k, and it is not cubic, so quoting P(2) = 6k and P(-2) = 8k as if they were determined is a genuine error even though their sum happens to be right. Choice (A) 0 agrees with 14k only in the degenerate case k = 0, where all three given values collapse to 0; the answer is asked in terms of k, so the general expression 14k is the one wanted. Note also that the argument never used A ≠ 0 — degree at most 3 is the whole hypothesis — so the answer is stable whether or not one insists a cubic have a genuine x³ term.
💡Key takeaway

When the inputs come in plus-and-minus pairs, add the polynomial to its mirror image — the odd terms cancel, and the coefficients you could never find cancel with them.

  • Name the four coefficients
  • Add P(x) and P(-x)
  • Read 2D and 2B off the givens
  • Evaluate the identity at x = 2
  • Build an actual cubic and confirm

More like this

Same archetype — closest grade level first.