AMC 10 · 2007 · #18

Grade 11 algebra
complex-numberspolynomial-rootspolynomial-factoringfunction-evaluation work-backwardspolynomial-factoring ↑ Prerequisites: complex-numberspolynomial-roots
📏 Long solution 💡 3 insights
Problem
A monic quartic with real coefficients has two given non-real roots. Find the sum of its four remaining coefficients.

Pick an answer.

(A)
0
(B)
1
(C)
4
(D)
9
(E)
16
How to solve
Strategy Work Backwards

The finish line is handed to us and the start is hidden: we know where f vanishes and want its coefficients, which is Tool #11 (Work Backwards) exactly. Tool #4 (Introduce a Variable) names the two missing roots r and s so that 'real coefficients' can be turned into an equation about them instead of a slogan. The step the whole problem actually turns on is not finding the conjugates, which is routine; it is showing that four roots pin a monic quartic down uniquely. That needs the four numbers to be pairwise distinct, and it is worth noticing why: if the two given roots had been 2i and -2i, conjugation would have produced nothing new, f would not be determined, and the question would have no answer. Once f is pinned down, Tool #7 (Identify Subproblems) does the multiplication in the cheap order, pairing each root with its own conjugate so that two real quadratics appear and the product is visibly real. Tool #15 (Organize Information in More Ways) supplies the independent check: a + b + c + d = f(1) - 1, so the whole answer can be read off one evaluation of f at x = 1 using moduli, without ever expanding the polynomial.

1STEP 1

Real coefficients force conjugate roots

Real coefficients bring in the two conjugate roots.

f(z) = f(z) ⟹ f(2i)=0 → f(-2i)=0, f(2+i)=0 → f(2-i)=0
2STEP 2

Four distinct roots pin f down

Four roots and a matching degree pin the polynomial exactly.

P(x) = (x-2i)(x+2i)(x-(2+i))(x-(2-i)) ∣ f(x), deg P = deg f = 4, both monic ⟹ f = P
3STEP 3

Pair each root with its conjugate

Pairing conjugates gives two real quadratics.

(x-2i)(x+2i) = x²+4, (x-(2+i))(x-(2-i)) = x²-4x+5
4STEP 4

Multiply out and read the coefficients

Multiplying out and adding gives 9.

f(x) = x⁴ - 4x³ + 9x² - 16x + 20 ⟹ a+b+c+d = -4 + 9 - 16 + 20 = 9
5STEP 5

Confirm by evaluating at x = 1

Evaluating at one confirms 9 without expanding, choice (D).

a+b+c+d = f(1)-1 = (1-2i)(1+2i)(-1-i)(-1+i) - 1 = 5 · 2 - 1 = 9
Answer
9
The strongest check is to verify the recovered polynomial against the original hypotheses rather than to re-add the coefficients. Take f(x) = x⁴ - 4x³ + 9x² - 16x + 20 and substitute x = 2i: the powers are (2i)⁴ = 16, (2i)³ = -8i, (2i)² = -4, so f(2i) = 16 + 32i - 36 - 32i + 20, whose real part is 16 - 36 + 20 = 0 and imaginary part is 32 - 32 = 0. Now substitute x = 2+i: successive powers are (2+i)² = 3+4i, (2+i)³ = 2+11i, (2+i)⁴ = -7+24i, giving real part -7 - 8 + 27 - 32 + 20 = 0 and imaginary part 24 - 44 + 36 - 16 = 0. Both hypotheses hold, and all four coefficients are real, so this polynomial really is admissible; Step 2 already showed it is the only admissible one, so 9 is forced. The distractors are all near misses from this same computation. Choice (E) 16 is |c|, what a solver reports after losing the sign on the x coefficient; choice (C) 4 is |a|, the same slip one term over; choice (A) 0 is a + c + d = -4 - 16 + 20, the total when b is skipped — a coincidence worth flagging, since it means the answer 9 happens to equal b alone; and choice (B) 1 is the leading coefficient, exactly the 1 that must be subtracted in the f(1) - 1 route and which is easy to leave in or take out at the wrong moment. A rough magnitude check also fits: d is the product of all four roots, |2i|²|2+i|² = 4 · 5 = 20, matching the constant term found by expansion.
💡Key takeaway

Real coefficients make roots come in mirror pairs, so two given roots become four; four different roots fill a quartic completely, and then x=1 hands you the sum of the coefficients in one shot.

  • Real coefficients force conjugate roots
  • Four distinct roots pin f down
  • Pair each root with its conjugate
  • Multiply out and read the coefficients
  • Confirm by evaluating at x = 1