Complex roots are always in pairs, and the pairs are conjugates. Therefore, even though you are only given two roots, the polynomial actually has three roots. The third root is #3 - i#.

Remember that a root is represented by #k#, and that the factor which yields a root is in the form #x - k#. Therefore, to write the polynomial which has the given roots and a leading coefficient of #1#, simply set up the roots in factor form and multiply them.

#f(x) = (x - -1)(x - [3 + i])(x - [3 - i])#

#f(x) = (x + 1)(x - 3 - i)(x - 3 + i)#

#f(x) = (x + 1)(x^2 - 3x +ix - 3x + 9 - 3i - ix + 3i - i^2)#

#f(x) = (x + 1)(x^2 - 6x + 9 +1)#

#f(x) = (x + 1)(x^2 - 6x + 10)#

#f(x) = x^3 - 6x^2 + 10x + x^2 - 6x + 10#

#f(x) = x^3 - 5x^2 + 4x + 10#