A polynomial with real coefficients the imaginary roots are conjigates of one another. So the roots are i, -i and 3 so the equation

f(x) = a(x-i)(x+i)(x-3)=a(x^2+1)(x-3) = a(x^3 - 3x^2 + x - 3)

I think you want f(2) = -10 so

f(2) = a(8 - 3(4) + 2 - 3) = a(8 - 12 + 2 - 3 = -5a so -5a = -10 so a = 2

so

`f(x) = 2x^3 - 6x^2 + 2x - 6`

Just to check

f(2) = 2(8) - 6(4) + 2(2) - 6 = 16 - 24 + 4 - 6 = -10

So our answer is

`f(x) = 2x^3 - 6x^2 + 2x - 6`