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 - i3−i.
Remember that a root is represented by kk, and that the factor which yields a root is in the form x - kx−k. Therefore, to write the polynomial which has the given roots and a leading coefficient of 11, 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 - 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)(x2−3x+ix−3x+9−3i−ix+3i−i2)
f(x) = (x + 1)(x^2 - 6x + 9 +1)f(x)=(x+1)(x2−6x+9+1)
f(x) = (x + 1)(x^2 - 6x + 10)f(x)=(x+1)(x2−6x+10)
f(x) = x^3 - 6x^2 + 10x + x^2 - 6x + 10f(x)=x3−6x2+10x+x2−6x+10
f(x) = x^3 - 5x^2 + 4x + 10f(x)=x3−5x2+4x+10
