Kako faktor x ^ 4 + 2x ^ 3y-3x ^ 2y ^ 2-4xy ^ 3-y ^ 4?

Kako faktor x ^ 4 + 2x ^ 3y-3x ^ 2y ^ 2-4xy ^ 3-y ^ 4?
Anonim

Odgovor:

# (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 #

Obrazloženje:

# "Riješite karakterističnu kvartičku jednadžbu bez y-ovog prvog:" #

# x ^ 4 + 2 x ^ 3 - 3 x ^ 2 - 4 x - 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 #

# "Ako to primijenimo na zadani polinom dobivamo" # #

# (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 #

# "(*) Uz zamjenu" x = y-1/2 "dobivamo:" #

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

# "Sada stavite" z = y ^ 2 "i pomnožite s 16:" #

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

# "disk:" 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 (21:00 4 sqrt (5)) / 2 #

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

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

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