Assuming that your question was stating that the polynomial has roots of -3 and #i# (the imaginary unit) as zeros, we can proceed as follows.
All complex number roots of the form #a+bi# of a polynomial are always accompanied by their complex conjugate, or the complex number #a-bi#. Because of this, we know that #-i# must also be a zero of #f(x)#, since #i# was a zero of #f(x)#.
We now have all three zeros of the function: -3, #i#, and #-i#. Expressing these as factors, we can then multiply out the factors to derive one such polynomial:
#f(x) = (x+3)(x-i)(x+i)#
# = (x+3)(x^2+ix-ix-i^2) = (x+3)(x^2-i^2) #
# = (x+3)(x^2+1) = (x^3+x+3x^2+3) #
# = x^3+3x^2+x+3 #
We can also verify that #f(1) = 8# as requested:
#f(1) = (1)^3 + 3(1)^2 + (1) + 3 = 1 + 3 + 1 + 3 = 8#
