Antwoord:
# (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 #
Uitleg:
# "Los de karakteristieke quartische vergelijking zonder de y's eerst op:" #
# 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 #
# "Als we dit toepassen op de gegeven veelterm krijgen we" #
# (x ^ 2 - x y - y ^ 2) (x ^ 2 + 3 xy + y ^ 2) = 0 #
# => (x- (1 + sqrt (5)) y / 2) (x- (1-vaks (5)) y / 2) #
# (X + (3 + sqrt (5)) y / 2) (x- (sqrt (5) -3) y / 2) = 0 #
# "(*) Met de vervanging" x = y-1/2 "krijgen we:" #
# y ^ 4 - (9/2) y ^ 2 + 1/16 = 0 #
# "Zet nu" z = y ^ 2 "en vermenigvuldig met 16:" #
# 16 z ^ 2 - 72 z + 1 = 0 #
# "schijf:" 72 ^ 2 - 4 * 16 = 5120 = 32 ^ 2 * 5 #
# => z = (72 uur 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 pm sqrt (5)) / 2 "of" (-3 pm sqrt (5)) / 2 #
