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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 14"

(Created page with "==Problem== Suppose that <math>P(z), Q(z)</math>, and <math>R(z)</math> are polynomials with real coefficients, having degrees <math>2</math>, <math>3</math>, and <math>6</mat...")
 
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==Solution==
 
==Solution==
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The answer can not be <math>0</math>, as every polynomial has at least <math>1</math> distinct complex root (fundamental theorem of algebra).
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The answer cannot be <math>0,</math> as every nonconstant polynomial has at least <math>1</math> distinct complex root (Fundamental Theorem of Algebra). Since <math>P(z) \cdot Q(z)</math> has degree <math>2 + 3 = 5,</math> we conclude that <math>R(z) - P(z)\cdot Q(z)</math> has degree <math>6</math> and is thus nonconstant.
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Take <math>P(z)=z^2+1,Q(z)=z^3+2,</math> and <math>R(z)=(z+1)^6 + \left(z^2+1\right)\left(z^3+2\right).</math>
 
  
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<math>R(z)</math> has degree 6 and constant term <math>3</math>, so it satisfies the conditions.
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It now suffices to illustrate an example for which <math>N = 1</math>: Take
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We need to find the solutions to <cmath>\left(z^2+1\right)\left(z^3+2\right)=(z+1)^6 + \left(z^2+1\right)\left(z^3+2\right),</cmath> or <cmath>(z+1)^6=0.</cmath> Clearly, there is one distinct complex root, <math>-1</math>, so our answer is <math>\boxed{\textbf{(B)} \: 1}.</math>
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<cmath>\begin{align*}
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P(z)&=z^2+1, \\
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Q(z)&=z^3+2, \\
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R(z)&=(z+1)^6 + P(z) \cdot Q(z).
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\end{align*}</cmath>
 +
Note that <math>R(z)</math> has degree <math>6</math> and constant term <math>3,</math> so it satisfies the conditions.
  
 +
We need to find the solutions to
 +
<cmath>\begin{align*}
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P(z) \cdot Q(z) &= (z+1)^6 + P(z) \cdot Q(z) \\
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0 &= (z+1)^6.
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\end{align*}</cmath>
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Clearly, the only distinct complex root is <math>-1,</math> so our answer is <math>N=\boxed{\textbf{(B)} \: 1}.</math>
  
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~kingofpineapplz
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~kingofpineapplz ~kgator
 +
 
 +
==Video Solution==
 +
https://youtu.be/HLhetkPGfX4
 +
 
 +
~MathProblemSolvingSkills.com
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2021 Fall|ab=B|num-a=13|num-b=11}}
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{{AMC12 box|year=2021 Fall|ab=B|num-a=15|num-b=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:47, 3 September 2022

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

Video Solution

https://youtu.be/HLhetkPGfX4

~MathProblemSolvingSkills.com

See Also

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