The polynomial is ASSUMED to be with real coefficients - it is the condition you missed in your post (!)


Then the roots are 5, -6, 1+i and 1-i  (together with 1+i, its conjugate 1-i is the root, too - for such a polynomial).


Then the associate linear binomials are 

    (x-5),  (x+6),  (x-(1+i))  and  (x-)1-i)).


We seek for the product of these binomials with the unknown real coefficient "a"

    f(x) = a*(x-5)*(x+6)*(x-(1+i))*(x-(1-i)).


The product of the last two binomials is

    (x-(1+i))*(x-(1-i)) = x^2 - 2x + (1+i)*(1-i) = x^2 - 2x +2.


Therefore, the polynomial f(x) is

     f(x) = a*(x-5)*(x+6)*(x^2 - 2x + 2)


To find "a", use the fact that  f(1) = 84.   It gives

      84 = a*(1-5)*(1+6)*(1 - 2 + 2) = a*(-4)*7*1 = -28a.


Hence,  a= 84/(-28) = -4.


So, your polynomial is

    f(x) = -4*(x-5)*(x+6)*(x^2 - 2x + 2).


It is your ANSWER.  


You may transform it further, if you want / (if you need).