Clearly the #x=-i# it satisfies the roots #x=+-i#
Thus we have #(x+1)^2#. Now using the other root (x+9) we can write a 3rd order polynomial...
#P_2=(x+1)^2 = x^2+2x+1 #
#P_3=P_2 (x+9) = x^3+11x^2+19x+9#
Remark: Notice that I did not use the root given in the question #x=-I#. The reason is that the complex number in general comes in pair out of even powered polynomial. For example out of the binomial. Recall the quadratic formula:
#x_(1,2) = (-b+- sqrt(b-4ac))/(2a)# Note that the complex number emerges for negative value of #(b-4ac) i.e. => b< 4ac# and note the true set is a pair of #+- " complex numbers"#
