Hi

finding all zeros of the function. 

 h(x)=x^3+4x^2+x-6  

      x^3+4x^2+x-6  = 0  |If x=1 then zero (1 + 4 + 1 - 6 = 0 )

   (x -1)(x+3)(x+2) =0  |Dividing  x^3+4x^2+x-6  by (x-1)

       (x-1)=0  x= 1

       (x+3)=0  x= -3

      (x+2) =0  x = -2

 

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