Check the picture below on the left-hand-side.
we know the degree of the polynomial will be 6, so the exponents of its factors must add up to 6.
now, we know that at -4 and -1 it has an odd multiplicity, so the exponent of that factor could be either 3 or 1 hmmm so which is it?
well, when the graph cross the axis straight through, it'd be a multiplicity of 1, when it leans a bit on the axis, like it does on -4, the multiplicity will be larger than 1, so then
[tex]\begin{cases} x=-4\implies &x+4=0\\ x=-1\implies &x+1=0\\ x=3\implies &x-3=0 \end{cases}\qquad \implies a\stackrel{odd\qquad odd\qquad even}{(x+4)^3(x+1)^1(x-3)^2}=y \\\\\\ \stackrel{\textit{we also know that}}{x=0\qquad y=576}\implies a(0+4)^3(0+1)(0-3)^2=576\implies a(64)(1)(9)=576 \\\\\\ 576a=576\implies a=\cfrac{576}{576}\implies a=1~\hfill \begin{array}{llll} 1\stackrel{odd\qquad odd\qquad even}{(x+4)^3(x+1)^1(x-3)^2}=y \\\\\\ \boxed{(x+4)^3(x+1)^1(x-3)^2=y} \end{array}[/tex]
Check the picture below on the right-hand-side
