Theorem. Let $f,g$ be polynomials. Suppose $f=Q_1g+R_1=Q_2g+R_2$, $deg(R_1)<deg(g)$ or $R_1=0$, and $deg(R_2)<deg(g)$ or $R_2=0$. Then $Q_1=Q_2$ and $R_1=R_2$.
$Proof$.
We have $Q_1g+R_1=Q_2g+R_2$, so $g(Q_1-Q_2)=R_2-R_1$. Suppose $Q_1-Q_2\neq0$ and $R_2-R_1\neq0$. Then $deg(Q_1-Q_2)\ge0$ and $deg(R_2-R_1)\ge0$. Now $$deg(g(Q_1-Q_2))=deg(g)+deg(Q_1-Q_2)=deg(R_2-R_1)\le max(deg(R_2),deg(R_1))<deg(g),$$ which is a contradiction. Therefore $Q_1=Q_2$ or $R_1=R_2$. In either case, the other equality can be derived from $Q_1g+R_1=Q_2g+R_2$.
Is the proof valid?