Question 729707
The two complex roots with imaginary components require two additional complex roots, the conjugates.


Along with (x-2), (x-(2+i)) makes (x-(2-i)) also necessary.   
(x-(1-i)) makes  (x-(1+i)) also necessary.  


The complex pairs can be turned into quadratic factors this way:
{{{(x-(2+i)) * (x-(2-i))=x^2-4x+5}}};
{{{(x-(1-i)) * (x-(1+i))=x^2-2x+2}}}.


A function with the five expected roots may be
{{{highlight(p(x)=(x-2)(x^2-4x+5)(x^2-2x+2))}}}.


Your next objective was to use that function as a starting relation to find a function to the degree 5 (?) which contains the point (0, -60).  You may possibly need a constant factor with the polynomial to be able to have the function value equal to -60.  Letting x=0 is simple; do this and then carry out the remaining multiplications.


{{{-60=(0-2)(0-0+5)(0-0+2)*k}}}, where k is some constant factor in case the original factors do not give -60.
By intent, I have not finished this work but it should be very easy.