Polynomial question, any way to solve for x?? URGENT!

Besides wikipedia,
http://simonyau2000.net/mathspace/algebra.pdf P.7-P.11
introduces two methods to solve a quartic equation.
They are: Descartes's formula and Ferrari's formula.

Descartes's formula is the most “user-friendly”. It is because Descartes's formula is the simplest and its idea is the most straight-forward. Ferrari's formula is too tedious.
連超級數學勁人冷凝液在http://hk.knowledge.yahoo.com/question/?qid=7006091400696都用Descartes's formula去解一元四次方程，所以我極力推薦用Descartes's formula去解一元四次方程。

Descartes's formula是用這樣的步驟去解一元四次方程Ax^4 + Bx³ + Cx² + Dx + E = 0。（經本人整理）

(1) 步驟一：將方程轉為化約(reduced)方程。

設x = y – h
則A(y – h)^4 + B(y – h)³ + C(y – h)² + D(y – h) + E = 0
The term in y³ will vanish if – 4Ah + B = 0
Choosing h = B/(4A), the equation becomes
A(y – B/(4A))^4 + B(y – B/(4A))³ + C(y – B/(4A))² + D(y – B/(4A)) + E = 0
A[y^4 – (B/A)y³ + (3B²/(8A²))y² – (B³/(16A³))y + B^4/(256A^4)] + B[y³ – (3B/(4A))y² + (3B²/(16A²))y – B³/(64A³)] + C[y² – (B/(2A))y + B²/(16A²)] + Dy – BD/(4A) + E = 0
Ay^4 + [3B²/(8A) – 3B²/(4A) + C]y² + [B³/(16A²) – 3B³/(16A²) – BC/(2A) + D]y + [B^4/(256A³) – B^4/(64A³) + B²C/(16A²) – BD/(4A) + E] = 0
Ay^4 + [(8AC – 3B²)/(8A)]y² + [(13B³ – 8ABC + 16A²D)/(16A²)]y + [(16(AB²C – 4A²BD + 16A³E) – 3B^4)/(256A³)] = 0
y^4 + [(8AC – 3B²)/(8A²)]y² + [(13B³ – 8ABC + 16A²D)/(16A³)]y + [(16(AB²C – 4A²BD + 16A³E) – 3B^4)/(256A^4)] = 0
設a = (8AC – 3B²)/(8A²)，b = (13B³ – 8ABC + 16A²D)/(16A³)及c = (16(AB²C – 4A²BD + 16A³E) – 3B^4)/(256A^4)，則方程變成
y^4 + ay² + by + c = 0。
這稱為化約方程。

(2) 步驟二：檢驗b是否為0。若b = 0，則可輕易求解。

若b = 0，則方程為(y²)² + ay² + c = 0，
即一個以y²為變元的一元二次方程，可由二次方程的求根公式解之：
y² = [– a ± √(a² – 4c)]/2
即 y = ± √{[– a ± √(a² – 4c)]/2} 或 ± √{[– a ± √(a² – 4c)]/2}
所以x = – B/(4A) ± √{[– a + √(a² – 4c)]/2} 或 – B/(4A) ± √{[– a – √(a² – 4c)]/2}

(3) 步驟三：若b ≠ 0，則需求解一個輔助方程（三次方程）。

設b ≠ 0。
由於任意一元四次實係數多項式均可寫成兩個一元二次實係數多項式的積，
所以可設y^4 + ay² + by + c ≡ (y² + ky + r)(y² – ky + s)，其中k、r和s是實數。
展開之，得y^4 + ay² + by + c ≡ y^4 + (r + s – k²)y² + k(s – r)y + rs。
由待定係數法，得r + s – k² = a，k(s – r) = b，及rs = c。
由b ≠ 0 得k ≠ 0。
由r + s = a + k²及s – r = b/k，得
2r = a + k² – b/k及2s = a + k² + b/k
所以4c = 4rs = (a + k² – b/k)(a + k² + b/k)
4c = (a + k²)² – b²/k²
4ck² = [(a + k²)k]² – b²
4ck² = k^6 + 2ak^4 + a²k² – b²
故得k^6 + 2ak^4 + (a² – 4c)k² – b² = 0
設F(z) = z³ + 2az² + (a² – 4c)z – b²。
則F(k²) = 0是一個以k²為變元的一元三次方程。
由於F(0) = – b² < 0，而當z足夠大時，F(z) > 0。
所以F(k²) = 0必有一正根。
解F(k²) = 0，得k的值。（最少有兩個實根，可任取其一）
至於如果解F(k²) = 0，
因為涉及一元三次方程，
不是本題的討論範圍內。
請自行前往http://hk.knowledge.yahoo.com/question/?qid=7007021400116，
果度會教你如何解一元三次方程。

(4) 步驟四：將方程化約為兩個一元二次方程並求解。

由於k已知，則r和s可由
r = (a + k² – b/k)/2及s = (a + k² + b/k)/2求得。
即y^4 + ay² + by + c = 0可化約為兩個二次方程：
y² + ky + r = 0及y² – ky + s = 0。
解之，得y = [– k ± √(k² – 4r)]/2 或 [k ± √(k² – 4s)]/2。
所以，原方程的解為
x = [– B – 2kA ± 2A√(k² – 4r)]/(4A) 或 [– B + 2kA ± 2A√(k² – 4s)]/(4A)。

2007-03-17 02:56:54 補充：
evilinyourheart, true, use newton's method to solve the quartic equation is also a kind of method.For the quartic equation Ax^4 Bx³ Cx² Dx E = 0,

2007-03-17 02:57:36 補充：
the iteration form using newton's method isx_(n 1) = x_n – (A(x_n)^4 B(x_n)³ C(x_n)² D(x_n) E)/(4A(x_n)³ 3B(x_n)² 2C(x_n) D).However, this method can ONLY find REAL roots, but not COMPLEX roots.So, this method does not quite good.

2007-03-17 03:08:11 補充：
evilinyourheart, true, use Newton's method to solve the quartic equation is also a kind of method.For the quartic equation Ax^4 十 Bx³ 十 Cx² 十 Dx 十 E = 0,

if you think that the method used above is too complicated,

try to use newton's method,although it uses up a lot of time.

it only requires the basic knowledge of calculus

let
f(x) = Ax^4 + Bx^3 + Cx^2 + Dx + E= 0

f'(x) = 4Ax^3 + 3Bx^2 + 2Cx + D

by

x[1] = x[0] - { f(x[0])/f'(x[1]) }

u can solve it

2007-03-08 17:05:05 補充：
x[1] = x[0] - { f(x[0])/f'(x[0]) }

Read this website for reference:
http://en.wikipedia.org/wiki/Quartic_equation
Your equation is a quartic equation.