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Algebra Difficulty 4.2 AIME Find the answer United States

Problem:

Find a real, irreducible quartic polynomial with leading coefficient 1 whose roots are all twelfth roots of unity.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

All twelfth roots of unity are roots of
x12−1=(x6−1)(x6+1)=(x3−1)(x3+1)(x6+1)=(x−1)(x2+x+1)(x+1)(x2−x+1)(x2+1)(x4−x2+1) \begin{aligned} x^{12}-1 & =\left(x^{6}-1\right)\left(x^{6}+1\right) \\ & =\left(x^{3}-1\right)\left(x^{3}+1\right)\left(x^{6}+1\right) \\ & =(x-1)\left(x^{2}+x+1\right)(x+1)\left(x^{2}-x+1\right)\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right) \end{aligned}
so the answer is x4−x2+1x^{4}-x^{2}+1.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.