# 五次方程

$y=x^5+13x^4+35x^3-85x^2-216x+252$的圖形

$ax^5+bx^4+cx^3+dx^2+ex+f=0,\,$

$x^5-4x^4+2x^3-3x+7=0\,$

## 布靈·傑拉德正規式

$x^5+a_1x^4+a_2x^3+a_3x^2+a_4x+a_5=0\,$

$y = x^4+b_1x^3+b_2x^2+b_3x+b_4\,$

$y^5 + my+ n=0\,$

$x= y-\frac{a_1}{5} \,$

$y^5+ay^3+by^2+cy+d=0\,$

$a=\frac{5a_2-2a_1^2}{5}\,$
$b=\frac{25a_3-15a_1a_2+4a_1^3}{25}\,$
$c=\frac{125a_4-50a_1a_3+15a_1^2a_2-3a_1^4}{125}\,$
$d=\frac{3125a_5-625a_1a_4+125a_1^2a_3-25a_1^3a_2+4a_1^5}{3125} \,$

$z^5+Pz^4+Qz^3+Az^2+Bz+C=0\,$

$p=\frac{-15b+\sqrt{60a^3+225b^2-200ac}}{10a} \,$
$q={2a \over 5} .\,$

$X=z^4+b_1z^3+b_2z^2+b_3z+b_4\,$

$z^5+Az^2+Bz+C=0\,$

$X^5+RX^4+SX^3+TX^2+UX+V=0\,$

$(27A^4-160B^3+300ABC) b_1^2+(27A^3B-400B^2C+375C^2A) b_1\,+(18A^2B^2-45A^3C-250BC^2)=0 ; \,$

$675A^3b_3^3+(3375A^2Cb_1-3600AB^2b_1-2025A^4\,-4500ABC)b_3^2$
$+(675A^3Bb_1^2+6000B^2Cb_1^2\,+7200A^2B^2b_1-4050A^3Cb_1+15000C^2Bb_1\,\!+9375C^3+9675A^2BC+2025A^5)b_3+\,$
$(-4770A^3BC-1125A^2C^2b_1^2-1500B^2C^2\,-320AB^3b_1^3-960B^4b_1^2-3843A^3B^2b_1+1485A^4Cb_1-54A^5b_1^3$
$-6250AC^3-2400B^3Cb_1-108A^2B^3-675A^6\,-756A^4Bb_1^2-9375AC^2Bb_1-3900AB^2Cb_1^2-225A^2BCb_1^3) = 0 .\,$

$X^5+UX+V= 0\,$

$X= \sqrt[4]{-U}\xi\,$

$\xi^5 - \xi+ t = 0\,$

## 特殊五次方程的求根公式

### 型式1

$ax^5+cx^3+\frac{c^2}{5}(-1)^kx+f=0\,$ $(a{(-1)^k}>0 \and k \in Z)\,$

$x_1=A+B,$
$x_2=c_1A+c_2B,$
$x_3=c_3A+c_4B,$
$x_4=c_2A+c_3B,$
$x_5=c_4A+c_1B,$

$A=\sqrt[5]{-\frac{f}{2a}+\sqrt{\frac{f^2}{4a^2}+\frac{c^5}{3125a^5}}}$
$B=\sqrt[5]{-\frac{f}{2a}-\sqrt{\frac{f^2}{4a^2}+\frac{c^5}{3125a^5}}}$
$c_1=\frac{(-1+\sqrt5)-\sqrt{10+2\sqrt5}{\mathrm{i}}}{4}$
$c_2=\frac{(-1+\sqrt5)+\sqrt{10+2\sqrt5}{\mathrm{i}}}{4}$
$c_3=\frac{(-1+\sqrt5)+\sqrt{10-2\sqrt5}{\mathrm{i}}}{4}$
$c_4=\frac{(-1+\sqrt5)-\sqrt{10-2\sqrt5}{\mathrm{i}}}{4}$

### 型式2

${}_{ax^5+bx^4+cx^3+\frac{15abc-4b^3}{25a^2}x^2+\frac{25a^2c^2-5ab^2c-b^4}{125a^3}x+f=0 ,\qquad\mbox{where }a\ne0}.$
${}_{x_1=\frac{-2b+\sqrt[5]{176b^5-1200ab^3c+2000a^2bc^2-50000a^4f+16\sqrt{\left(11b^5-75ab^3c+125a^2bc^2-3125a^4f\right)^2-4\left(2b^2-5ac\right)^5}}+\sqrt[5]{176b^5-1200ab^3c+2000a^2bc^2-50000a^4f-16\sqrt{\left(11b^5-75ab^3c+125a^2bc^2-3125a^4f\right)^2-4\left(2b^2-5ac\right)^5}}}{10a}}\,$
${}_{x_2=-\frac{b}{5a}+\frac{-1+\sqrt5+\sqrt{10+2\sqrt5}{\rm{i}}}{40a}\sqrt[5]{176b^5-1200ab^3c+2000a^2bc^2-50000a^4f+16\sqrt{\left(11b^5-75ab^3c+125a^2bc^2-3125a^4f\right)^2-4\left(2b^2-5ac\right)^5}}+\frac{-1+\sqrt5-\sqrt{10+2\sqrt5}{\rm{i}}}{40a}\sqrt[5]{176b^5-1200ab^3c+2000a^2bc^2-50000a^4f-16\sqrt{\left(11b^5-75ab^3c+125a^2bc^2-3125a^4f\right)^2-4\left(2b^2-5ac\right)^5}}}\,$
${}_{x_3=-\frac{b}{5a}+\frac{-1-\sqrt5+\sqrt{10-2\sqrt5}{\rm{i}}}{40a}\sqrt[5]{176b^5-1200ab^3c+2000a^2bc^2-50000a^4f+16\sqrt{\left(11b^5-75ab^3c+125a^2bc^2-3125a^4f\right)^2-4\left(2b^2-5ac\right)^5}}+\frac{-1-\sqrt5-\sqrt{10-2\sqrt5}{\rm{i}}}{40a}\sqrt[5]{176b^5-1200ab^3c+2000a^2bc^2-50000a^4f-16\sqrt{\left(11b^5-75ab^3c+125a^2bc^2-3125a^4f\right)^2-4\left(2b^2-5ac\right)^5}}}\,$
${}_{x_4=-\frac{b}{5a}+\frac{-1-\sqrt5-\sqrt{10-2\sqrt5}{\rm{i}}}{40a}\sqrt[5]{176b^5-1200ab^3c+2000a^2bc^2-50000a^4f+16\sqrt{\left(11b^5-75ab^3c+125a^2bc^2-3125a^4f\right)^2-4\left(2b^2-5ac\right)^5}}+\frac{-1-\sqrt5+\sqrt{10-2\sqrt5}{\rm{i}}}{40a}\sqrt[5]{176b^5-1200ab^3c+2000a^2bc^2-50000a^4f-16\sqrt{\left(11b^5-75ab^3c+125a^2bc^2-3125a^4f\right)^2-4\left(2b^2-5ac\right)^5}}}\,$
${}_{x_5=-\frac{b}{5a}+\frac{-1+\sqrt5-\sqrt{10+2\sqrt5}{\rm{i}}}{40a}\sqrt[5]{176b^5-1200ab^3c+2000a^2bc^2-50000a^4f+16\sqrt{\left(11b^5-75ab^3c+125a^2bc^2-3125a^4f\right)^2-4\left(2b^2-5ac\right)^5}}+\frac{-1+\sqrt5+\sqrt{10+2\sqrt5}{\rm{i}}}{40a}\sqrt[5]{176b^5-1200ab^3c+2000a^2bc^2-50000a^4f-16\sqrt{\left(11b^5-75ab^3c+125a^2bc^2-3125a^4f\right)^2-4\left(2b^2-5ac\right)^5}}}\,$

### 型式3

$(c^2+1)x^5+5d^4(3\mp c)x-4d^5(\pm 11+2c)=0,\qquad\mbox{where }c\ne \pm{\rm{i}}\,$
$x_1=d\left[A+B+C+D\right]\,$
$x_2=d\left[\frac{(-1+\sqrt5)+\sqrt{10+2\sqrt5}{\rm{i}}}{4}A+\frac{(-1-\sqrt5)+\sqrt{10-2\sqrt5}{\mathrm{i}}}{4}B\right]\,$
$+d\left[\frac{(-1-\sqrt5)-\sqrt{10-2\sqrt5}{\mathrm{i}}}{4}C+\frac{(-1+\sqrt5)-\sqrt{10+2\sqrt5}{\mathrm{i}}}{4}D\right]\,$
$x_3=d\left[\frac{(-1-\sqrt5)+\sqrt{10-2\sqrt5}{\mathrm{i}}}{4}A+\frac{(-1-\sqrt5)-\sqrt{10-2\sqrt5}{\mathrm{i}}}{4}B\right]\,$
$+d\left[\frac{(-1+\sqrt5)-\sqrt{10+2\sqrt5}{\mathrm{i}}}{4}C+\frac{(-1+\sqrt5)+\sqrt{10+2\sqrt5}{\mathrm{i}}}{4}D\right]\,$
$x_4=d\left[\frac{ (-1-\sqrt5)-\sqrt{10-2\sqrt5}{\mathrm{i}}}{4}A+\frac{(-1+\sqrt5)-\sqrt{10+2\sqrt5}{\mathrm{i}}}{4}B\right]\,$
$+d\left[\frac{(-1+\sqrt5)+\sqrt{10+2\sqrt5}{\mathrm{i}}}{4}C+\frac{(-1-\sqrt5)+\sqrt{10-2\sqrt5}{\mathrm{i}}}{4}D\right]\,$
$x_5=d\left[\frac{(-1+\sqrt5)-\sqrt{10+2\sqrt5}{\rm{i}}}{4}A+\frac{(-1+\sqrt5)+\sqrt{10+2\sqrt5}{\mathrm{i}}}{4}B\right]\,$
$+d\left[\frac{(-1-\sqrt5)+\sqrt{10-2\sqrt5}{\mathrm{i}}}{4}C+\frac{(-1-\sqrt5)-\sqrt{10-2\sqrt5}{\mathrm{i}}}{4}D\right]\,$

$A=\sqrt[5]{\frac{\left( \sqrt{c^2+1} +\sqrt{c^2+1\mp\sqrt{c^2+1}}\right)^2\left(-\sqrt{c^2+1}+\sqrt{c^2+1\pm\sqrt{c^2+1}}\right)} {(c^2+1)^2}}\,$
$B=\sqrt[5]{\frac{\left( \sqrt{c^2+1} +\sqrt{c^2+1\mp\sqrt{c^2+1}}\right)^2\left(-\sqrt{c^2+1}+\sqrt{c^2+1\pm\sqrt{c^2+1}}\right)} {(c^2+1)^2}}\,$
$C=\sqrt[5]{\frac{\left( \sqrt{c^2+1} +\sqrt{c^2+1\pm\sqrt{c^2+1}}\right)^2\left(\sqrt{c^2+1}-\sqrt{c^2+1\mp\sqrt{c^2+1}}\right)} {(c^2+1)^2}}\,$
$D=-\sqrt[5]{\frac{\left( \sqrt{c^2+1} -\sqrt{c^2+1\mp\sqrt{c^2+1}}\right)^2\left(-\sqrt{c^2+1}+\sqrt{c^2+1\pm\sqrt{c^2+1}}\right)} {(c^2+1)^2}}\,$

### 型式4

$a^2x^5+5abx^3+5b^2x+c=0,\qquad\mbox{where }a\ne 0\,$

$x_1=A+B\,$

$x_2=\frac{(-1+\sqrt5)+\sqrt{10+2\sqrt5}{\rm{i}}}{4}A+\frac{(-1+\sqrt5)-\sqrt{10+2\sqrt5}{\rm{i}}}{4}B\,$

$x_3=\frac{(-1-\sqrt5)+\sqrt{10-2\sqrt5}{\rm{i}}}{4}A+\frac{(-1-\sqrt5)-\sqrt{10-2\sqrt5}{\rm{i}}}{4}B\,$

$x_4=\frac{(-1+\sqrt5)-\sqrt{10+2\sqrt5}{\rm{i}}}{4}A+\frac{(-1+\sqrt5)+\sqrt{10+2\sqrt5}{\rm{i}}}{4}B\,$

$x_5=\frac{(-1-\sqrt5)-\sqrt{10-2\sqrt5}{\rm{i}}}{4}A+\frac{(-1-\sqrt5)+\sqrt{10-2\sqrt5}{\rm{i}}}{4}B\,$

$A=\sqrt[5]{-\frac{c}{2a}+\sqrt{\left(\frac{c}{2a}\right)^2+\left(\frac{b}{a}\right)^5}}\,$ $B=\sqrt[5]{-\frac{c}{2a}-\sqrt{\left(\frac{c}{2a}\right)^2+\left(\frac{b}{a}\right)^5}}\,$

## 引文

1. ^ 阿米爾·艾克塞爾（Amir D. Aezel）. 費馬最後定理. 台北: 時報出版. 1998: p.87. ISBN 957-13-2648-8.