AMC 10 · 2021 · #15

Grade 11 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Symmetric Function of Roots via Vieta
vieta-formulascomplex-numberspolynomial-rootssymmetric-polynomials identify-subproblemspattern-recognition ↑ Prerequisites: vieta-formulascomplex-numbers
📏 Long solution 💡 3 insights
Problem
A quartic has four complex roots. Each root is fed through a map that conjugates it and then multiplies by four i. The four outputs are the roots of a new monic quartic. Find the sum of that new quartic's quadratic coefficient and its constant term.

Pick an answer.

(A)
${-}304$
(B)
${-}208$
(C)
12i
(D)
208
(E)
304
How to solve
Strategy Change Focus / Count the Complement

Hunting for the four roots of P is hopeless: it is a quartic with no obvious factorization. So change focus. A polynomial's coefficients are not built from any single root; they are built from sums and products that use all four roots at once and do not care about their order. Those combinations are exactly what Vieta's formulas hand us for free from the coefficients of P. Since B and D are two such combinations of the new roots, and each new root is the old root conjugated and scaled by the same fixed number 4i, the whole computation can be done on the combinations without ever knowing a single root.

1STEP 1

Read the root sums off P

Read the symmetric sums off the original.

Σ z_i = -4, Σ_i ＜ j z_i z_j = 3, Σ_i ＜ j ＜ k z_i z_j z_k = -2, z₁ z₂ z₃ z₄ = 1
2STEP 2

Say what B and D really are

Say what the two coefficients really are.

B = Σ_i ＜ j w_i w_j, D = w₁ w₂ w₃ w₄
3STEP 3

Pull the factor 4i outside

Pull the constant factor outside.

Σ_i ＜ j w_i w_j = (4i)² Σ_i ＜ j z_i z_j, w₁ w₂ w₃ w₄ = (4i)⁴ z₁ z₂ z₃ z₄
4STEP 4

Conjugation changes nothing here

With real coefficients, conjugation changes nothing.

Σ_i ＜ j z_i z_j = Σ_i ＜ j z_i z_j = 3 = 3, z₁ z₂ z₃ z₄ = 1 = 1
5STEP 5

Multiply out and add

Multiplying and adding gives 208.

B = (4i)² · 3 = -48, D = (4i)⁴ · 1 = 256, B + D = 208
Answer
208
The constant term is an easy independent check: the product of the four new roots is (4i)⁴ times the product of the four old roots, and (4i)⁴ = 256 while the old product is 1, so D = 256 — a positive real number, as it must be since i⁴ = 1. That already forces the answer above 200, so -304 and -208 are out, and B = -48 pulls the total down from 256 to 208 rather than up to 304. Choice (C) 12i can be rejected without any computation: the roots of P come in conjugate pairs, so the new roots do too up to the fixed factor, and the symmetric combinations stay real. B + D = 208 is consistent with all of it.
💡Key takeaway

You do not need the roots to know a polynomial's coefficients — the coefficients are built from sums and products of all the roots at once, so scaling every root by 4i just multiplies those packages by powers of 4i.

  • Read the root sums off P
  • Say what B and D really are
  • Pull the factor 4i outside
  • Conjugation changes nothing here
  • Multiply out and add

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