SOLUTION: Let {{{g(x)= x^4-4x^3-4x^2-4x-6 }}} Given that the complex number i is a zero of the function, find all the zeros of the function.

x=i is not a zero of the polynomial above.  So you must have
accidently copied it wrong.  If the -6 on the end were a -5 
instead, then i would be a zero.  So I'm going to assume you 
meant -5 on the end instead of -6. 

Since x=i is a zero, its conjugate x=-i is also a zero.

Since x=i is equivalent to x-i=0 and x=-i is equivalent to x+i=0

Therefore x-i and x+i are both factors of g(x)

Therefore (x-i)(x+i) is a factor of g(x)

(x-i)(x+i) = x2-i2 = x2-(-1) = x2+1 = x2+0x+1

So we divide by long division:

                x²-4x-5
x²+0x+1)x⁴-4x³-4x²-4x-5 
        x⁴-0x³+ x² 
          -4x³-5x²-4x
          -4x³-0x²-4x
              -5x²+0x-5
              -5x²-0x-5
                      0

So g(x) has now been factored as

   g(x) = (x²+1)(x²-4x-5)

Then we factor the quadratic expression in the second
parentheses:

   g(x) = (x²+1)(x-5)(x+1)

So the zeros are

          x²+1=0;  x-5=0;   x+1=0
            x²=-1;   x=5;     x=-1
             x=±i;

The zeros are i, -i, 5 and -1
  
Edwin