-1, 2, 3i; and f(-2)=10

Since 3i is a solution, its conjugate -3i is also a solution.
[Note that since 3i is really 0+3i, its conjugate is 0-3i, which
is -3i]

So we start with

 x = -1;   x = 2,   x = 3i,  x = -3i

Get 0 on the right side of each:

  
x+1 = 0;   x-2 = 0  x-3i = 0,  x+3i = 0

Myltiply all the left and right sides together.  The right side
will just be 0:

   (x+1)(x-2)(x-3i)(x+3i) = 0

Multiply both sides by a constant k, and the left
side will be f(x) if we have the right value for k

f(x) = k(x+1)(x-2)(x-3i)(x+3i) 

Multiply the first two and the last two factors  

f(x) = k(x²-2x+1x-2)(x²+3ix-3ix-9i²) 

f(x) = k(x²-x-2)(x²-9i²) 

Now since i² = -1, -9i² = -9(-1) = +9

f(x) = k(x²-x-2)(x²+9) 

Multiply those two parentheses together:

f(x) = k(x4+9x²-x³-2x²-9x-18) 
   
Collect like terms:

f(x) = k(x4-x³+7x²-9x-18)

Now we can find k because we are given f(-2)=10

We substitute x=-2

f(-2) = k((-2)4-(-2)³+7(-2)²-9(-2)-18)

f(-2) = k(16-(-8)+7(4)+18-18)

f(-2) = k(16+8+28)

f(-2) = k(52)

We substitute 10 for f(-2)

  10 = k(52)

 = k       

 = k

Substitute for k in:

f(x) = k(x4-x³+7x²-9x-18)

f(x) = (x4-x³+7x²-9x-18)

f(x) = x4 - x³ + x² - x - 

Edwin