AMC 12A 2022 (Problem 21)Let P(x)=x2022+x1011+1P(x)=x^{2022}+x^{1011}+1P(x)=x2022+x1011+1. Which of the following polynomials is a factor of P(x)P(x)P(x)?(A) x2−x+1\text{(A)}\;x^2-x+1(A)x2−x+1(B) x2+x+1\text{(B)}\;x^2+x+1(B)x2+x+1(C) x4+1\text{(C)}\;x^4+1(C)x4+1(D) x6−x3+1\text{(D)}\;x^6-x^3+1(D)x6−x3+1(E) x6+x3+1\text{(E)}\;x^6+x^3+1(E)x6+x3+1Related TopicsCoreToolkit 32 — Remainder of Polynomial DivisionHints (5)Hint 1Checking Choice (A): x2−x+1=0x^2-x+1=0x2−x+1=0.x2=x−1x^2=x-1x2=x−1x3=x(x−1)=x2−x=x−1−x=−1x^3=x(x-1)=x^2-x=x-1-x=-1x3=x(x−1)=x2−x=x−1−x=−1P(x)=(x3)674+(x3)337+1P(x)=(x^3)^{674}+(x^3)^{337}+1P(x)=(x3)674+(x3)337+1=(−1)674+(−1)337+1=1−1+1=1≠0.=(-1)^{674}+(-1)^{337}+1=1-1+1=1\ne0.=(−1)674+(−1)337+1=1−1+1=1=0.Hint 2Checking Choice (B): x2+x+1=0x^2+x+1=0x2+x+1=0.x2=−x−1x^2=-x-1x2=−x−1x3=x(−x−1)=−x2−x=−(−x−1)−x=x+1−x=1x^3=x(-x-1)=-x^2-x=-(-x-1)-x=x+1-x=1x3=x(−x−1)=−x2−x=−(−x−1)−x=x+1−x=1P(x)=(x3)674+(x3)337+1P(x)=(x^3)^{674}+(x^3)^{337}+1P(x)=(x3)674+(x3)337+1=1674+1337+1=3≠0.=1^{674}+1^{337}+1=3\ne0.=1674+1337+1=3=0.Hint 3Checking Choice (C): x4+1=0x^4+1=0x4+1=0.x4=−1x^4=-1x4=−1P(x)=(x4)505x2+(x4)252x3+1P(x)=(x^4)^{505}x^2+(x^4)^{252}x^3+1P(x)=(x4)505x2+(x4)252x3+1=(−1)505x2+(−1)252x3+1=−x2+x3+1≠0.=(-1)^{505}x^2+(-1)^{252}x^3+1=-x^2+x^3+1\ne0.=(−1)505x2+(−1)252x3+1=−x2+x3+1=0.Hint 4Checking Choice (D): x6−x3+1=0x^6-x^3+1=0x6−x3+1=0.x6=x3−1x^6=x^3-1x6=x3−1x9=x3x6=x3(x3−1)=x6−x3=x3−1−x3=−1x^9=x^3x^6=x^3(x^3-1)=x^6-x^3=x^3-1-x^3=-1x9=x3x6=x3(x3−1)=x6−x3=x3−1−x3=−1P(x)=(x9)224x6+(x9)112x3+1P(x)=(x^9)^{224}x^6+(x^9)^{112}x^3+1P(x)=(x9)224x6+(x9)112x3+1=(−1)224x6+(−1)112x3+1=x6+x3+1=(-1)^{224}x^6+(-1)^{112}x^3+1=x^6+x^3+1=(−1)224x6+(−1)112x3+1=x6+x3+1=(x3−1)+x3+1=2x3≠0.=(x^3-1)+x^3+1=2x^3\ne0.=(x3−1)+x3+1=2x3=0.Hint 5Checking Choice (E): x6+x3+1=0x^6+x^3+1=0x6+x3+1=0.x6=−x3−1x^6=-x^3-1x6=−x3−1x9=x3x6=x3(−x3−1)=−x6−x3=x3+1−x3=1x^9=x^3x^6=x^3(-x^3-1)=-x^6-x^3=x^3+1-x^3=1x9=x3x6=x3(−x3−1)=−x6−x3=x3+1−x3=1P(x)=(x9)224x6+(x9)112x3+1P(x)=(x^9)^{224}x^6+(x^9)^{112}x^3+1P(x)=(x9)224x6+(x9)112x3+1=1224x6+1112x3+1=x6+x3+1=0.=1^{224}x^6+1^{112}x^3+1=x^6+x^3+1=0.=1224x6+1112x3+1=x6+x3+1=0.Final Answer(E) x6+x3+1x^6+x^3+1x6+x3+1Related Problems (2)AMC 10B 2022 (Problem 21)AMC 12B 2021 Spring (Problem 20)