Hi

finding all zeros of the function. 

 f(x)= x^3-5x^2-18x +72

  x^3-5x^2-18x +72 = 0  |If x=3 then zero (27 -45 -54 + 72 = 0)

   (x -3)(x-6)(x+4) =0  |Dividing x^3-5x^2-18x +72  by (x-3)

  (x-3)=0  x = 3

  (x-6)=0  x = 6

  (x+4)=0  x = -4 

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