Question 792243
Recheck your problem description.  If P is degree two, then you have TWO zeros; you cannot have zeros 1+i, 1-i, and 5 as the zeros.  IF you have 1+i as a zero, then you ALSO have 1-i as a zero because complex zeros occur in P as "Conjugate Pairs", so you cannot also have 5 as a zero.  If your zeros are supposed to be 1+i and 5, then you must also have 1-i; and this means P must be AT LEAST degree 3.  


For degree THREE, your question would work and have a solution, but not for degree TWO.  


You would, for degree THREE, have factors for P being {{{(x-5)}}}, {{{(x-(1+i))}}}, and {{{(x-(1-i))}}}.


Recheck your problem description from your book or source carefully, and then ask what you really want help with.