2021 Fall AMC 12B Problems/Problem 14

Problem

Suppose that $P(z), Q(z)$ , and $R(z)$ are polynomials with real coefficients, having degrees $2$ , $3$ , and $6$ , respectively, and constant terms $1$ , $2$ , and $3$ , respectively. Let $N$ be the number of distinct complex numbers $z$ that satisfy the equation $P(z) \cdot Q(z)=R(z)$ . What is the minimum possible value of $N$ ?

$\textbf{(A)}\: 0\qquad\textbf{(B)} \: 1\qquad\textbf{(C)} \: 2\qquad\textbf{(D)} \: 3\qquad\textbf{(E)} \: 5$

Solution

The answer cannot be $0,$ as every nonconstant polynomial has at least $1$ distinct complex root (Fundamental Theorem of Algebra). Since $P(z) \cdot Q(z)$ has degree $2 + 3 = 5,$ we conclude that $R(z) - P(z)\cdot Q(z)$ has degree $6$ and is thus nonconstant.

It now suffices to illustrate an example for which $N = 1$ : Take $\begin{align*} P(z)&=z^2+1, \\ Q(z)&=z^3+2, \\ R(z)&=(z+1)^6 + P(z) \cdot Q(z). \end{align*}$ Note that $R(z)$ has degree $6$ and constant term $3,$ so it satisfies the conditions.

We need to find the solutions to $\begin{align*} P(z) \cdot Q(z) &= (z+1)^6 + P(z) \cdot Q(z) \\ 0 &= (z+1)^6. \end{align*}$ Clearly, the only distinct complex root is $-1,$ so our answer is $N=\boxed{\textbf{(B)} \: 1}.$

~kingofpineapplz ~kgator

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2021 Fall AMC 12B Problems/Problem 14

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