□投稿者/ Fa 一般人(1回)-(2007/07/21(Sat) 18:31:07)
■No26555に返信(らすかるさんの記事)
> もし問題が x^4+5x+9 で正しいとしたら、途中計算は省略しますが、答は
> t=2√{(√3)cos(arccos(225√3/1296)/3)} とおいて
> {2x+t+i√(t^2-10/t)}{2x+t-i√(t^2-10/t)}{2x-t+i√(t^2+10/t)}{2x-t-i√(t^2+10/t)}/16

　　　　　　　　拡大体で　（因数）困数ぶんかい　　；
(-(1/2)*Sqrt[12/(1/2*(25 + I*Sqrt[6287]))^(1/3) + (1/2*(25 + I*Sqrt[6287]))^(1/3)] -
1/2*Sqrt[-(12/(1/2*(25 + I*Sqrt[6287]))^(1/3)) - (1/2*(25 + I*Sqrt[6287]))^
(1/3) - 10/Sqrt[12/(1/2*(25 + I*Sqrt[6287]))^(1/3) + (1/2*(25 +
I*Sqrt[6287]))^(1/3)]] + x)*
(-(1/2)*Sqrt[12/(1/2*(25 + I*Sqrt[6287]))^(1/3) + (1/2*(25 + I*Sqrt[6287]))^
(1/3)] +
1/2*Sqrt[-(12/(1/2*(25 + I*Sqrt[6287]))^(1/3)) - (1/2*(25 + I*Sqrt[6287]))^
(1/3) - 10/Sqrt[
12/(1/2*(25 + I*Sqrt[6287]))^(1/3) + (1/2*(25 + I*Sqrt[6287]))^(1/3)]] + x)*
(1/2*Sqrt[12/(1/2*(25 + I*Sqrt[6287]))^(1/3) + (1/2*(25 + I*Sqrt[6287]))^
(1/3)] -
1/2*Sqrt[-(12/(1/2*(25 + I*Sqrt[6287]))^(1/3)) - (1/2*(25 + I*Sqrt[6287]))^
(1/3) + 10/Sqrt[
12/(1/2*(25 + I*Sqrt[6287]))^(1/3) + (1/2*(25 +
I*Sqrt[6287]))^(1/3)]] + x)*
(1/2*Sqrt[12/(1/2*(25 + I*Sqrt[6287]))^(1/3) + (1/2*(25 + I*Sqrt[6287]))^(1/3)] +
1/2*Sqrt[-(12/(1/2*(25 + I*Sqrt[6287]))^(1/3)) - (1/2*(25 + I*Sqrt[6287]))^
(1/3) + 10/Sqrt[12/(1/2*(25 + I*Sqrt[6287]))^(1/3) + (1/2*(25 +
I*Sqrt[6287]))^(1/3)]] + x)