Hvordan faktor du helt: x ^ 8-9?

Svar:

# X ^ 8-9 = (x-3 ^ (1/4)) (x + 3 ^ (1/4)) (x-i3 ^ (1/4)) (x + i3 ^ (1/4)) (x- (1 / sqrt (2) + i / sqrt (2)) 3 ^ (1/4)) (x + (1 / sqrt (2) + i / sqrt (2)) 3 ^ (1/4)) (x- (1 / sqrt (2) -i / sqrt (2)) 3 ^ (1/4)) (x + (1 / sqrt (2) -i / sqrt (2)) 3 ^ (1 / 4)) #

Forklaring:

Brug af forskellen i kvadratfaktorisering (# A ^ 2-b ^ 2 = (a-b) (a + b) #) du har:

# X ^ 8-9 = (x ^ 4-3) (x ^ 4 + 3) #

Dette er nok alt, hvad de vil, men du kan faktor yderligere tillade komplekse tal:

# (X ^ 4-3) (x ^ 4 + 3) = #

# (X ^ 2-3 ^ (1/2)) (x ^ 2 + 3 ^ (1/2)) (x ^ 2-i3 ^ (1/2)) (x ^ 2 + i3 ^ (1 / 2)) = #

# (X-3 ^ (1/4)) (x + 3 ^ (1/4)) (x-i3 ^ (1/4)) (x + i3 ^ (1/4)) (x- (1 / sqrt (2) + i / sqrt (2)) 3 ^ (1/4)) (x + (1 / sqrt (2) + i / sqrt (2)) 3 ^ (1/4)) (x- (1 / sqrt (2) -i / sqrt (2)) 3 ^ (1/4)) (x + (1 / sqrt (2) -i / sqrt (2)) 3 ^ (1/4)) #

De 8 rødder er de 8 løsninger til: # X ^ 8 = 9 #

Svar:

faktor # x ^ 8 - 9 #

Forklaring:

# x ^ 8 - 9 = (x ^ 4 - 3) (x ^ 4 + 3) = #

= # (x ^ 2 - sqrt3) (x ^ 2 + sqrt3) (x ^ 4 + 3) #

= # x-rod (4) (3)) (x + root (4) (3)) (x ^ 2 + sqrt3) (x ^ 4 + 3) #