Find all real and complex zeroes of x^3-4x^2+24

x^3-4x^2+24  = 0     

We could use the Rational Zeros Theorem to find at least one real root.......however.....in this case, a graph is faster

https://www.desmos.com/calculator/wgw3soov8t

There is only one real root at  (-2, 0)

Using synthetic division, we have

-2     [  1       - 4          0          24  ]

                     -2        12         -24

         ------------------------------------

            1       -6         12          0

This tells us that the remaining polynomial set to 0 is

x^2 - 6x  + 12  =  0    complete the square

x^2 - 6x  + 9  =  -12 + 9

(x - 3)^2   = -3         take the pos/neg    roots

x - 3  =  pos/neg sqrt (-3)

x - 3   pos/neg sqrt(3)i

x = pos/neg sqrt(3)i + 3