Question 419258
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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