上級者
数学Ⅰ：方程式と不等式
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ヒント：\(x^{2n}+x^n+1\)を直接因数分解するのは難しい．ここで，\(X=x^n\)とすると，\(X^2+X+1\)の形は\(X^3-1\)\(=(X^2+X+1)\)\((X-1)\)となり，\(X^3-1\)と\(X-1\)を因数分解すればよい．\(x^n-1=0\)の解は複素数の範囲で見たことがあると思います．\(x\)を直接求めると複素数範囲ですが，複素数平面に解を置くと対称性が見えてきます．それをうまく使うと実数範囲の因数を求めることができます．
問題
(1)\(n\)を自然数とする．\(x^{2n}+x^n+1\)を係数が実数の範囲で因数分解しなさい．
(2)\(x^{10}+x^{5}+1\)を係数が整数の範囲で因数分解しなさい．
解答
(1)
\((x^{2n}+x^n+1)\)\((x^n-1)\)\(=(x^{3n}-1)\)となる．
ここで，複素数平面を考えると，
\(\displaystyle w_{k}=\cos\frac{2k\pi}{3n}+i\sin\frac{2k\pi}{3n}\)と置くとき，
\(x^n-1\)\(=(x-w_{0})\)\((x-w_{3})\)\(\cdots\)\((x-w_{3(n-1)})\)
\(x^{3n}-1\)\(=(x-w_{0})\)\((x-w_{1})\)\(\cdots\)\((x-w_{3n-1})\)
ここで，
\(w_{k}+w_{3n-k}\)\(=\displaystyle \cos\frac{2k\pi}{3n}+i\sin\frac{2k\pi}{3n}\)\(+\displaystyle \cos\frac{(3n-2k)\pi}{3n}+i\sin\frac{(3n-2k)\pi}{3n}\)\(=\displaystyle \cos\frac{2k\pi}{3n}+i\sin\frac{2k\pi}{3n}\)\(+\displaystyle \cos\frac{2k\pi}{3n}-i\sin\frac{2k\pi}{3n}\)\(=\displaystyle 2\cos\frac{2k\pi}{3n}\)
\(w_{k}w_{3n-k}\)\(=\displaystyle \cos\frac{(2k+3n-2k)\pi}{3n}+i\sin\frac{(2k+3n-2k)\pi}{3n}\)\(=1\)
ここで，\(n\)が偶数の時，\(n=2m\)と置くと，
\(x^{6m}-1\)\(=(x-w_{1})\)\((x-w_{6m-1})\)\((x-w_{2})\)\((x-w_{6m-2})\)\(\cdots\)\((x-w_{3m-1})\)\((x-w_{3m+1})\)\((x-w_{0})\)\((x-w_{3m})\)\(=(x^2-2\cos\frac{2\pi}{6m}x+1)\)\((x^2-2\cos\frac{4\pi}{6m}x+1)\)\(\cdots\)\((x^2-2\cos\frac{2(3m-1)\pi}{6m}x+1)\)\((x-w_{0})\)\((x-w_{3m})\)
\(x^{2m}-1\)\(=(x-w_{3})\)\((x-w_{6m-3})\)\((x-w_{6})\)\((x-w_{6m-6})\)\(\cdots\)\((x-w_{3(m-1)})\)\((x-w_{3(m+1)})\)\((x-w_{0})\)\((x-w_{3m})\)\(=(x^2-2\cos\frac{6\pi}{6m}x+1)\)\((x^2-2\cos\frac{12\pi}{6m}x+1)\)\(\cdots\)\((x^2-2\cos\frac{6(m-1)\pi}{6m}x+1)\)\((x-w_{0})\)\((x-w_{3m})\)
よって，\(x^{2n}+x^n+1\)\(=(x^2-2\cos\frac{\pi}{3m}x+1)\)\((x^2-2\cos\frac{2\pi}{6m}x+1)\)\((x^2-2\cos\frac{4\pi}{6m}x+1)\)\((x^2-2\cos\frac{5\pi}{6m}x+1)\)\(\cdots\)\((x^2-2\cos\frac{2(3m-2)\pi}{6m}x+1)\)\((x^2-2\cos\frac{2(3m-1)\pi}{6m}x+1)\)\(=(x^2-2\cos\frac{2\pi}{3n}x+1)\)\((x^2-2\cos\frac{4\pi}{3n}x+1)\)\((x^2-2\cos\frac{8\pi}{3n}x+1)\)\((x^2-2\cos\frac{10\pi}{3n}x+1)\)\(\cdots\)\((x^2-2\cos\frac{(3n-4)\pi}{3n}x+1)\)\((x^2-2\cos\frac{(3n-2)\pi}{3n}x+1)\)
ここで，\(n\)が奇数の時，\(n=2m-1\)と置くと，
\(x^{6m-3}-1\)\(=(x-w_{1})\)\((x-w_{6m-4})\)\((x-w_{2})\)\((x-w_{6m-5})\)\(\cdots\)\((x-w_{3m-2})\)\((x-w_{3m-1})\)\((x-w_{0})\)\(=(x^2-2\cos\frac{2\pi}{3(2m-1)}x+1)\)\((x^2-2\cos\frac{4\pi}{3(2m-1)}x+1)\)\(\cdots\)\((x^2-2\cos\frac{2(3m-2)\pi}{3(2m-1)}x+1)\)\((x-w_{0})\)
\(x^{2m-1}-1\)\(=(x-w_{3})\)\((x-w_{6m-6})\)\((x-w_{6})\)\((x-w_{6m-9})\)\(\cdots\)\((x-w_{3m-3})\)\((x-w_{3m})\)\((x-w_{0})\)\(=(x^2-2\cos\frac{6\pi}{3(2m-1)}x+1)\)\((x^2-2\cos\frac{12\pi}{3(2m-1)}x+1)\)\(\cdots\)\((x^2-2\cos\frac{2(3m-3)\pi}{3(2m-1)}x+1)\)\((x-w_{0})\)
よって，\(x^{2n}+x^n+1\)\(=(x^2-2\cos\frac{2\pi}{3(2m-1)}x+1)\)\((x^2-2\cos\frac{4\pi}{3(2m-1)}x+1)\)\((x^2-2\cos\frac{8\pi}{3(2m-1)}x+1)\)\((x^2-2\cos\frac{10\pi}{3(2m-1)}x+1)\)\(\cdots\)\((x^2-2\cos\frac{2(3m-4)\pi}{3(2m-1)}x+1)\)\((x^2-2\cos\frac{2(3m-2)\pi}{3(2m-1)}x+1)\)\(=(x^2-2\cos\frac{2\pi}{3n}x+1)\)\((x^2-2\cos\frac{4\pi}{3n}x+1)\)\((x^2-2\cos\frac{8\pi}{3n}x+1)\)\((x^2-2\cos\frac{10\pi}{3n}x+1)\)\(\cdots\)\((x^2-2\cos\frac{(3n-5)\pi}{3n}x+1)\)\((x^2-2\cos\frac{(3n-1)\pi}{3n}x+1)\)
(2)
係数が整数になる場合，\(\displaystyle 2\cos\frac{2k\pi}{3n}\)が整数となる場合を調べればよい．つまり，\(\displaystyle 2\cos\frac{2k\pi}{3n}=\pm 1\)\(\Leftrightarrow\)\(\displaystyle \frac{2k\pi}{3n}\)\(=\displaystyle \frac{\pi}{3}\),\(\displaystyle \frac{2\pi}{3}\)\(\Leftrightarrow\)\(\displaystyle k=\frac{n}{2}\),\(k=n\)（\(\displaystyle k\leqq \frac{3n-1}{2}\)で\(k\)は自然数）
ここで，\(x^{10}+x^{5}+1\)の因数は，\(n=5\)で\(x^2-2\cos\frac{20\pi}{15}x+1\)\((=x^2+x+1)\)となる．
よって，\(x^{10}+x^{5}+1\)を\(x^2+x+1\)で割ると，\(x^8-x^7+x^5-x^4+x^3-x+1\)となる．
よって，\(x^{10}+x^{5}+1\)\(=(x^2+x+1)\)\((x^8-x^7+x^5-x^4+x^3-x+1)\)
※直接解く場合は，
\(x^2+x+1=0\)の\(x\)の解を\(\omega_{1}\)，\(\omega_{2}\)と置く．\(x^{10}+x^{5}+1\)に\(x=\omega_{1}\)を代入するとき，\(\omega_{1}^2=-(1+\omega_{1})\)を使って変形していくと，
\(\omega_{1}^{10}+\omega_{1}^{5}+1\)\(=-(1+\omega_{1})^{5}+\omega_{1}^{5}+1\)\(=-5\omega_{1}(\omega_{1}^3+2\omega_{1}^2+2\omega_{1}+1)\)\(=-5\omega_{1}(\omega_{1}(\omega_{1}^2+\omega_{1}+1)+\omega_{1}^2+\omega_{1}+1)\)\(=0\)
また，\(\omega_{2}\)も同様なので，\(x^{10}+x^{5}+1\)は\(x^2+x^x+1\)を因数に持つ．後は割ってみる．
参考問題（第121回数学検定 1級1次）
\(x^{14}+x^7+1\)を係数が実数の範囲で因数分解しなさい．
解
\(x^{14}+x^7+1\)\(=\left(x^2-2\cos\frac{2\pi}{21}x+1\right)\)\(\left(x^2-2\cos\frac{4\pi}{21}x+1\right)\)\(\left(x^2-2\cos\frac{8\pi}{21}x+1\right)\)\(\left(x^2-2\cos\frac{10\pi}{21}x+1\right)\)\(\left(x^2-2\cos\frac{2\pi}{3}x+1\right)\)\(\left(x^2-2\cos\frac{16\pi}{21}x+1\right)\)\(\left(x^2-2\cos\frac{20\pi}{21}x+1\right)\)