Find all the zeros of 

f(x) = x5 - 10x4 + 42x3 -124x2 + 297x - 306 

given that one zero is 3i

We will use synthetic division to divide by (x - 3i)

3i | 1    -10      42      -124       297       -306
   |         +3i   -9-30i    90+99i  -297-102i   306 
     1    -10+3i   33-30i   -34+99i      -102i     0

Now we have factored the polynomial as

f(x) = (x - 3i)[x4 + (-10+3i)x³ + (33-30i)x² - (-102i)x + (-102)]

When a polynomial with all real coefficients has a complex imaginary zero,
its conjugate is also a zero.

Therefore -3i is also a zero

We will use synthetic division to divide

 x4 + (-10+3i)x³ + (33-30i)x² - (-102i)x + (-102) by (x - 3i)

-3i |  1 -10+3i 33-30i -34+99i -102i
    |       -3i    30i    -99i -102i 
       1 -10    33      -34        0

Now we have factored the polynomial as
f(x) = (x - 3i)(x + 3i)(x³ - 10x² + 33x - 34)

We now seek zeros of x³ - 10x² + 33x - 34

The possible rational zeros are ± factors of 34, so they are ±1, ±2, ±17, ± 34

We will use synthetic division to divide by "x - (each of those)" to see if any 
are zeros:

We try dividing by x - 1 to see if 1 is a zero:

1 | 1 -10 33 -34 
  |     1 -9  24 
    1  -9 24 -10

Nope. We didn't get a remainder of 0.

We try dividing by x + 1 to see if -1 is a zero:
-1 | 1 -10 33 -34 
   |    -1 11 -44 
     1 -11 44 -78

Nope. We didn't get a remainder of 0.

We try dividing by x - 2 to see if 2 is a zero:
2 | 1 -10  33 -34 
  |     2 -16  34 
    1  -8  17   0

We DID get 0 remainder, so now we have factored the
polynomial as

f(x) = (x - 3i)(x + 3i)(x - 2)(x² - 8x + 17)

The last parentheses contains an unfactorable quadratic,
So we use the quadratic formula:

x² - 8x + 17, with a=1, b=-8, c=17

x = 

x = 

x = 

x = 

x = 

x = 

x = 

x = 4 ± i 

So the zeros are 3i, -3i, 2, 4+i, 4-i

Edwin