2021 Fall AMC 12B Problems/Problem 14

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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 can not be $0$, as every polynomial has at least $1$ distinct complex root (fundamental theorem of algebra). Take $P(z)=z^2+1,Q(z)=z^3+2,$ and $R(z)=(z+1)^6 + \left(z^2+1\right)\left(z^3+2\right).$

$R(z)$ has degree 6 and constant term $3$, so it satisfies the conditions. We need to find the solutions to \[\left(z^2+1\right)\left(z^3+2\right)=(z+1)^6 + \left(z^2+1\right)\left(z^3+2\right),\] or \[(z+1)^6=0.\] Clearly, there is one distinct complex root, $-1$, so our answer is $\boxed{\textbf{(B)} \: 1}.$


~kingofpineapplz

See Also

2021 Fall AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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