Question 419257
  <pre><font face = "Tohoma" size = 4 color = "indigo"><b>
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