Bagaimana Anda memfaktorkan x ^ 4 + 2x ^ 3y-3x ^ 2y ^ 2-4xy ^ 3-y ^ 4?

Bagaimana Anda memfaktorkan x ^ 4 + 2x ^ 3y-3x ^ 2y ^ 2-4xy ^ 3-y ^ 4?
Anonim

Menjawab:

# (x- (1 + sqrt (5)) y / 2) (x- (1-sqrt (5)) y / 2) #

# (x + (3 + sqrt (5)) y / 2) (x- (sqrt (5) -3) y / 2) = 0 #

Penjelasan:

# "Selesaikan persamaan kuartik karakteristik tanpa y yang pertama:" #

# x ^ 4 + 2 x ^ 3 - 3 x ^ 2 - 4x - 1 = 0 #

# => (x ^ 2-x-1) (x ^ 2 + 3x + 1) = 0 "(*)" #

# "1)" x ^ 2 + 3x + 1 = 0 => x = (-3 pm sqrt (5)) / 2 #

# "2)" x ^ 2-x-1 = 0 => x = (1 pm sqrt (5)) / 2 #

# "Jika kita menerapkan ini pada polinomial yang diberikan, kita mendapat" #

# (x ^ 2 - x y - y ^ 2) (x ^ 2 + 3 xy + y ^ 2) = 0 #

# => (x- (1 + sqrt (5)) y / 2) (x- (1-sqrt (5)) y / 2) #

# (x + (3 + sqrt (5)) y / 2) (x- (sqrt (5) -3) y / 2) = 0 #

# "(*) Dengan substitusi" x = y-1/2 "kita mendapatkan:" #

# y ^ 4 - (9/2) y ^ 2 + 1/16 = 0 #

# "Sekarang letakkan" z = y ^ 2 "dan kalikan dengan 16:" #

# 16 z ^ 2 - 72 z + 1 = 0 #

# "disc:" 72 ^ 2 - 4 * 16 = 5120 = 32 ^ 2 * 5 #

# => z = (72 pm 32 sqrt (5)) / 32 = 9/4 pm sqrt (5) #

# => y = pm sqrt (9/4 pm sqrt (5)) #

# => y = pm sqrt (9 pm 4 sqrt (5)) / 2 #

# => y = pm sqrt ((2 pm sqrt (5)) ^ 2) / 2 #

# => y = pm (1 pm sqrt (5) / 2) #

# => x = (1 siang sqrt (5)) / 2 "atau" (-3 pm sqrt (5)) / 2 #