SOLUTION: The problem is as follows= Given that x=1+2i is a zero for the funtion g(x)=3x^3-2x^2+7x+20, determine the remaining zeros. The only solutions I know are 1+2i and 1-2i but I am

g(x) = 3x³ - 2x² + 7x + 20

First divide synthetically by [x - (1+2i)]

    1+2i | 3 -2      7     20
         |    3+6i -11+8i -20
           3  1+6i  -4+8i   0

Now we have partially factored the polynomial g(x) this way:

g(x) = [x - (1+2i)][3x² + (1+6i)x + (-4+8i)]

Now we divide the second factor synthetically by [x - (1-2i)]

    1-2i | 3  1+6i  -4+8i   
         |    3-6i   4-8i
           3  4      0

Now we have completely factored the polynomial g(x) this way:

g(x) = [x - (1+2i)][x - (1-2i)](3x + 4)

Setting the first factor = 0 gives x = 1+2i, which you were given.
Setting the second factor = 0 gives x = 1-2i, which you knew was
a zero because it is the conjugate of 1+2i.
Setting the third factor = 0,

      3x + 4 = 0
          3x = -4
           x = 

That's the third zero. 
           
Edwin