2019 AMC 12A Problems/Problem 14

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 $|c| = \boxed{\textbf{(E) } \sqrt{10}}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2019 AMC 12A Problems/Problem 14

Problem

Solution

See Also