Answer :
One way to do is to write the first polynomial in terms of the second; this means find [tex]a,b,c,d[/tex] so that
[tex]3x^3+5x^2-48x-80=a(x+4)^3+b(x+4)^2+c(x+4)+d[/tex]
Expanding the right side and matching up coefficients of terms with the same power of [tex]x[/tex] gives
[tex]\begin{cases}a=3\\12a+b=5\\48a+8b+c=-48\\64a+16b+4c+d=-80\end{cases}\implies a=3,b=-31,c=56,d=0[/tex]
So we have
[tex]3x^3+5x^2-48x-80=3(x+4)^3-31(x+4)^2+56(x+4)[/tex]
and in particular we can see [tex]x+4[/tex] divides this exactly, giving us
[tex]\dfrac{3x^3+5x^2-48x-80}{x+4}=3(x+4)^2-31(x+4)+56[/tex]
and expanding gives
[tex]\dfrac{3x^3+5x^2-48x-80}{x+4}=\boxed{3x^2-7x-20}[/tex]