AMC 10 · 2021 · #14

Grade 11 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
polynomial-rootscomplex-numberspolynomial-factoringextremal-construction work-backwardsextreme-principle ↑ Prerequisites: polynomial-rootscomplex-numbers
📏 Medium solution 💡 3 insights
Problem
Three polynomials with real coefficients are given only by their degree and their constant term. Consider the complex solutions where the product of the first two equals the third. Among all legal choices, find the smallest possible number of distinct solutions.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
5
How to solve
Strategy Introduce a Variable

An equation between two unknown polynomials is hard to look at, but the difference of the two sides is a single polynomial, so I name it S(z) = P(z)Q(z) - R(z) and count its distinct roots instead. Degree and constant term are the only data given, and both behave predictably under multiplying and subtracting, so I can pin down exactly what S looks like. That gives a floor on N immediately. For the minimum I then work backwards: decide what the ideal S would be if all roots piled onto one point, and check that such an S really comes from a legal P, Q, R.

1STEP 1

Name the difference of the two sides

Name the difference of the two sides.

S(z) = P(z)Q(z) - R(z), P(z)Q(z) = R(z) ⇔ S(z) = 0
2STEP 2

Read off the degree and constant term of S

The largest degree decides the total.

deg(PQ) = 2 + 3 = 5 ＜ 6 → deg S = 6, S(0) = 1 · 2 - 3 = -1
3STEP 3

Zero roots is impossible

A positive degree means at least one root.

deg S = 6 ≥ 1 → S has a complex root → N ≥ 1
4STEP 4

Any degree-6 target with S(0)=-1 is legal

Any target respecting the constant term is legal.

P(z)=z²+1, Q(z)=z³+2, R(z)=P(z)Q(z)-S(z) → deg R = 6, R(0)=2-(-1)=3
5STEP 5

Pile all six roots onto one point

Piling all roots onto one point gives 1.

R(z)-P(z)Q(z) = (z-1)⁶ = 0 → z = 1 only → N_min = 1
Answer
1
The two halves of the argument meet exactly: N ≥ 1 holds for every legal choice, and one explicit choice reaches N = 1, so the minimum is 1 and nothing smaller is possible. The witness survives every check by hand: P(z)=z²+1 has degree 2 and constant term 1, Q(z)=z³+2 has degree 3 and constant term 2, and R(z) = (z²+1)(z³+2) + (z-1)⁶ = z⁶ + … + 3 has degree 6 and constant term 3, all with real coefficients. Note the constant terms are not decoration: they force S(0) = -1 ≠ 0, so z = 0 can never be the repeated root and a tempting shape like S(z) = -z⁶ is banned; shifting the root to z=1 fixes that. The answer is also the smallest choice above 0, which fits a problem whose only real content is ruling 0 out.
💡Key takeaway

Move everything to one side and look at the leftover polynomial: it can never be constant here, so at least one root always exists, and stacking all six roots on the same point makes that one root the whole answer.

  • Name the difference of the two sides
  • Read off the degree and constant term of S
  • Zero roots is impossible
  • Any degree-6 target with S(0)=-1 is legal
  • Pile all six roots onto one point

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