<s>[INST] Find every $a,b \in \mathbb{C}$ such that dividing $P=3X^{120}+\sqrt{2}X^{19}+aX^{11}+\sqrt{2}X^9+3X^2+b$ by $Q=X^2+1$ has remainder $R=5X+2$ We know that, for some $H\in\mathbb{C}[x]$,
$$P = QH+R$$
Hence,
$$P-R=QH$$
Given that $Q$'s roots are $\{i,-i\}$, the Remainder Theorem for polynomial rings guarantees that evaluating $(P-R)(\pm i)=0$. So we have the following system of equations
$$\begin{cases} 3i^{120}+\sqrt{2}i^{19}+ai^{11}+\sqrt{2}i^9+3i^2+5i+b+2 =0 \\ 3(-i)^{120}+\sqrt{2}(-i)^{19}+a(-i)^{11}+\sqrt{2}(-i)^9+3(-i)^2+5(-i)+b+2 =0 \end{cases}$$
which entails that 
$$a=-5-\sqrt{2},b=2$$
Is this more or less correct and properly justified?
 [/INST] Right method, but there are sign errors:
$$\begin{cases} 3i^{120}+\sqrt{2}i^{19}+ai^{11}+\sqrt{2}i^9+3i^2\color{red}-5i+b\color{red}-2 =0 \\ 3(-i)^{120}+\sqrt{2}(-i)^{19}+a(-i)^{11}+\sqrt{2}(-i)^9+3(-i)^2\color{red}-5(-i)+b\color{red}-2 =0 \end{cases} \Rightarrow \\
\begin{cases} \require{cancel}\cancel{3}-\cancel{\sqrt{2}i}-ai+\cancel{\sqrt{2}i}-\cancel{3}\color{red}-5i+b\color{red}-2 =0 \\ \cancel3+\cancel{\sqrt{2}i}+ai-\cancel{\sqrt{2}i}-\cancel{3}+5i+b\color{red}-2 =0 \end{cases} \Rightarrow a=-5,b=2.$$
Alternatively, note:
$$3x^{120}+3x^2\equiv 0\pmod{x^2+1}\\
\sqrt{2}x^{19}+\sqrt{2}x^9 \equiv 0\pmod{x^2+1}\\
ax^{11}+b\equiv 5x+2\pmod{x^2+1} \Rightarrow \\
ax^{11}+b=g(x)(x^2+1)+5x+2 \Rightarrow \\
ai^{11}+b=g(i)\cdot 0+5i+2\Rightarrow \\
-ai+b=5i+2 \Rightarrow \\
a=-5,b=2.$$
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