AMC 10 · 2009 · #21

Grade 11 algebra
Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type
polynomial-rootscomplex-numbersroots-of-unity polynomial-substitutioncasework ↑ Prerequisites: polynomial-roots
📏 Long solution 💡 3 insights
Problem
A cubic's coefficients are reused to build a degree-twelve polynomial. Count how many of its zeros are not real.

Pick an answer.

(A)
4
(B)
6
(C)
8
(D)
10
(E)
12
How to solve
Strategy Solve an Easier Related Problem

A degree-12 polynomial with unknown complex coefficients looks hopeless head-on. But the exponents 12, 8, 4, 0 are the exponents 3, 2, 1, 0 of p multiplied by 4, so the big polynomial is just p with x⁴ fed into it. That turns one hard degree-12 question into the easy cubic I already know everything about, plus three separate fourth-root questions. I first work backwards from the three given zeros to write p in factored form, then split the degree-12 zero set into three quartic families, then decide in each family which zeros are real. Two things the count really turns on and that I must check, not assume: that the three families do not overlap and that each family really has four different zeros; otherwise "count the zeros" and "count the distinct zeros" would disagree.

1STEP 1

Work backwards to factor p

The three zeros factor the cubic completely.

p(x) = (x - (2009 + 9002π i))(x - 2009)(x - 9002)
2STEP 2

Read the big polynomial as p(x⁴)

The big polynomial is that cubic with a fourth power inside.

x¹² + ax⁸ + bx⁴ + c = p(x⁴) = (x⁴ - (2009 + 9002π i))(x⁴ - 2009)(x⁴ - 9002)
3STEP 3

Check the twelve zeros are distinct

All twelve zeros are distinct.

x⁴ = r_i and x⁴ = r_j → r_i = r_j; r_i ≠ 0 → x⁴ - r_i has 4 distinct zeros
4STEP 4

Family x⁴ = 2009 + 9002 pi i

The non-real target gives four non-real zeros.

x ∈ R → x⁴ ∈ R, but Im(2009 + 9002π i) = 9002π ≠ 0 → 4 nonreal zeros
5STEP 5

Families x⁴ = 2009 and x⁴ = 9002

Each real target gives two more.

x⁴ - t = (x² - √(t))(x² + √(t)) → x = ±∜(t) (real), x = ± i∜(t) (nonreal)
6STEP 6

Tally, then cross-check by complement

Tallying and cross-checking gives 8, choice (B).

4 + 2 + 2 = 8 = 12 - 4
Answer
8
The count 8 sits inside the possible range 0 to 12 and is one of the listed choices. It also passes a structural test: the four real zeros ±∜(2009) and ±∜(9002) can be checked one at a time, since (±∜(2009))⁴ = 2009 makes the middle factor of p(x⁴) vanish, and likewise for 9002. One tempting shortcut must be avoided: nonreal zeros come in conjugate pairs only when the coefficients are real, and here they are not, because a = -(2009 + 9002π i) - 2009 - 9002 is itself nonreal. So the evenness of 8 is a coincidence of this problem, not a forced symmetry, and the count had to be built family by family. Finally, the giant-looking numbers 2009, 9002, and π never entered the arithmetic; only three facts about them mattered: two are positive reals, one is nonreal, and all three are different and nonzero.
💡Key takeaway

When the exponents of a polynomial are all multiples of 4, it is really a smaller polynomial with x⁴ plugged in, and each of its zeros splits into four fourth roots you can sort into real and nonreal.

  • Work backwards to factor p
  • Read the big polynomial as p(x⁴)
  • Check the twelve zeros are distinct
  • Family x⁴ = 2009 + 9002 pi i
  • Families x⁴ = 2009 and x⁴ = 9002
  • Tally, then cross-check by complement

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