Odgovor: Ničla od #f (x) = x ^ 4-6x ^ 2 + 8 # so # {sqrt2, -sqrt2,2, -2} # Pojasnilo: Najprej razčlenimo #f (x) = x ^ 4-6x ^ 2 + 8 # = # x ^ 4-4x ^ 2-2x ^ 2 + 8 # = # x ^ 2 (x ^ 2-4) -2 (x ^ 2-4) # = # (x ^ 2-2) (x ^ 2-4) # = # (x ^ 2- (sqrt2) ^ 2) (x ^ 2-2 ^ 2) # = # (x-sqrt2) (x + sqrt2) (x-2) (x + 2) # To pomeni za # x = {sqrt2, -sqrt2,2, -2} # imamo #f (x) = 0 # Zato so ničle #f (x) = x ^ 4-6x ^ 2 + 8 # so # {sqrt2, -sqrt2,2, -2} #
