<s>[INST] Polynomial remainder when divisor has double roots How to find the remainder of the division of $P(x) = x^{444} + x^{111} + x - 1$ by $Q(x) = (x^2 + 1)^2$?
By the little Bézout's theorem, we can write
$$P(x) = (x^2 + 1)^2 W(x) + ax^3+bx^2+cx+d.$$
In fact, I've found out that, since $(x^2 + 1)$ divides $P(x)$, we can write:
$$P(x) = (x^2 + 1)^2 W(x) + cx^3+dx^2+cx+d.$$
Next, I'd usually plug the roots of $(x^2 + 1)^2$, which are $\{-i,i\}$, into the preceding expression, to get a system of equations that would allow me to find the unknowns $a,b,c,d$ (or just $c,d$ in this case). But as each root has multiplicity 2, I end up with an indeterminate system. What am I missing?
 [/INST] We have
\begin{eqnarray}
P(x)&=&(x^2+1)^2W(x)+ax^3+bx^2+cx+d\\
P'(x)&=&4x(x^2+1)W(x)+(x^2+1)^2W'(x)+3ax^2+2bx+c
\end{eqnarray}
and
$$
P(i)=i^{444}+i^{111}+i-1=(i^4)^{111}+i^{108}i^3+i-1=1-i+i-1=0
$$
It follows that
$$
0=P(i)=ai^3+bi^2+ci+d=-ai-b+ci+d=(d-b)+(c-a)i
$$
Therefore
$$
a=c,\quad b=d
$$
Also
$$
P'(x)=444x^{443}-111x^{110}-1
$$
therefore
$$
P'(i)=444i^{443}-111i^{110}-1=444i^{440}i^3-111i^{108}i^2-1=110-444i
$$
Substituting $c=a$, we get
$$
110-444i=P'(i)=3ai^2+2bi+c=c-3a+2bi=-2a+2bi
$$
it follows that
$$
a-bi=-55+222i
$$
Therefore
$$
a=c=-55,\quad b=d=-222.
$$
Hence the remainder is 
$$
R(x)=-55x^3-222x^2-55x-222=-(55x+222)(x^2+1).
$$
</s>