Problem:
Let
P(x)=x2022+x1011+1
Which of the following polynomials is a factor of P(x) ?
Answer Choices:
A. x2βx+1
B. x2+x+1
C. x4+1
D. x6βx3+1
E. x6+x3+1
Solution:
Note that
P(x)=x1011β1x3033β1β.
Its roots are 3033 rd roots of unity that are not 1011th roots of unity. Because
x6+x3+1=x3β1x9β1β
the roots of choice (E) are the six different complex numbers z that are ninth roots but not cube roots of unity; that is, z9=1 and z3ξ =1. For any such z, observe that z3033=1 and z1011=z1008β
z3ξ =1, implying that P(z)=0. Therefore x6+x3+1 is a factor of P(x).
The roots of the other choices are not roots of P(x). The roots of choices (A) and (B) are sixth roots of unity, but not square roots:
(x2βx+1)(x2+x+1)=x2β1x6β1β.
The roots of choice (C) are eighth roots of unity, but not fourth roots:
(x4+1)=x4β1x8β1β.
The roots of choice (D) are eighteenth roots of unity, but not sixth or ninth roots:
(x6βx3+1)=(x6+x3+1)(x6β1)x18β1β.
The problems on this page are the property of the MAA's American Mathematics Competitions