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Difference between revisions of "2019 AMC 12A Problems/Problem 14"
 
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<math>P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)</math>
 
<math>P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)</math>
  
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by using the quadratic formula on the quadratic factors. Since the first four roots are all distinct, the term <math>(x^2 - cx + 4)</math> must be a product of any combination of two (not necessarily distinct) factors from the set: <math>(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),</math> and <math>(x - [2 + 2i])</math>. We need the two factors to yield a constant term of <math>4</math> when multiplied together. The only combinations that work are <math>(x - [1 - i])</math> and <math>(x - [2 + 2i])</math>, or <math>(x - [1+i])</math> and <math>(x - [2-2i])</math>. When multiplied together, the polynomial is either <math>(x^2 + [-3 + i]x + 4)</math> or <math>(x^2+[-3-i]x+4)</math>. Therefore, <math>c = 3 \pm i</math> and <math>|c| = \boxed{\textbf{(E) } \sqrt{10}}</math>.
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by using the [[quadratic formula]] on each of the quadratic factors. Since the first four roots are all distinct, the term <math>(x^2 - cx + 4)</math> must be a product of any combination of two (not necessarily distinct) factors from the set: <math>(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),</math> and <math>(x - [2 + 2i])</math>. We need the two factors to yield a constant term of <math>4</math> when multiplied together. The only combinations that work are <math>(x - [1 - i])</math> and <math>(x - [2 + 2i])</math>, or <math>(x - [1+i])</math> and <math>(x - [2-2i])</math>. When multiplied together, the polynomial is either <math>(x^2 + [-3 + i]x + 4)</math> or <math>(x^2+[-3-i]x+4)</math>. Therefore, <math>c = 3 \pm i</math> and <math>|c| = \boxed{\textbf{(E) } \sqrt{10}}</math>.
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2019|ab=A|num-b=13|num-a=15}}
 
{{AMC12 box|year=2019|ab=A|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Intermediate Algebra Problems]]

Latest revision as of 22:53, 25 July 2022

Problem

For a certain complex number $c$, the polynomial \[P(x) = (x^2 - 2x + 2)(x^2 - cx + 4)(x^2 - 4x + 8)\]has exactly 4 distinct roots. What is $|c|$?

$\textbf{(A) } 2 \qquad \textbf{(B) } \sqrt{6} \qquad \textbf{(C) } 2\sqrt{2} \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \sqrt{10}$

Solution

The polynomial can be factored further broken down into

$P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)$

by using the quadratic formula on each of the quadratic factors. Since the first four roots are all distinct, the term $(x^2 - cx + 4)$ must be a product of any combination of two (not necessarily distinct) factors from the set: $(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),$ and $(x - [2 + 2i])$. We need the two factors to yield a constant term of $4$ when multiplied together. The only combinations that work are $(x - [1 - i])$ and $(x - [2 + 2i])$, or $(x - [1+i])$ and $(x - [2-2i])$. When multiplied together, the polynomial is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \pm i$ and $|c| = \boxed{\textbf{(E) } \sqrt{10}}$.

See Also

2019 AMC 12A (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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