Question 1000165
(x+4)^2 deals with root of -4 with multiplicity 2.
(1/2a)(-b+/- sqrt(b^2-4ac) has to be -2 +5i and -2 -5i, because complex roots are conjugates.  The square root has to be -25 or some multiple of it. b^2-4ac=-25*constant.   b=2 or some multiple of it. Let's make a 1, (1/2)(4+/- sqrt(16-4c)). Here, b is 4. To get 5i, I need the square root of 100, because I want 10i, which divided by 2 gives 5i.  So 4c=116, and c=29.
x^2-4x+29 would be that polynomial.
(x+4)^2(x^-4x+29)
x^4+4x^3+13x^2+168x+464

{{{graph(300,200,-10,10,-50,1000,x^4+4x^3+13x^2+168x+464)}}}