A. Find the zeros of the function algebraically. (Enter your answers as a comma-separated list.) f(x) = 1/3x^3 - x B. Find the zeros of the function algebraically. Multiple part answer, please list steps. f(x) = 5x^2 + 23x - 10 To find the zeros, set the function equal to zero and factor. 0 = (___ x - ___) (x+ ___)... Solution A) f(x) = 1/3x^3 - x = x(1/3 x2 -1) zeroes of f(x) are: -sqrt3 , 0, sqrt 3 B)f(x) = 5x^2 + 23x - 10for zeroes set f(x)=0 5x^2 + 23x - 10=0 5x2 + 25x -2x -10 = 0 or, 5x (x+5) -2(x+5) = 0 or, (x+5)(5x-2)=0 or x=-5 , 2/5 zeroes of f(x) are : -5 , 2/5.