= Solution
All the following <polynomial coefficients> and sums lie in the <field with two elements>, so $a+a=0$. For the constant <polynomial>,
$$
a=(x+1)\cdot0+a.
$$
For the linear <polynomial>, subtracting the leading multiple means adding it:
$$
ax+b=(x+1)a+(a+b).
$$
For the quadratic <polynomial>, first cancel $ax^2$ using $ax(x+1)$, leaving $(a+b)x+c$; then cancel $(a+b)x$ using $(a+b)(x+1)$. Thus
$$
ax^2+bx+c=(x+1)\bigl(ax+a+b\bigr)+(a+b+c).
$$
The requested <polynomial quotients> and <polynomial remainders> are therefore
$$
\boxed{\begin{aligned}
a &: &&Q=0, &&R=a,\\
ax+b &: &&Q=a, &&R=a+b,\\
ax^2+bx+c &: &&Q=ax+a+b, &&R=a+b+c.
\end{aligned}}
$$
For a general <polynomial> $P(x)=\sum_{i=0}^{n}p_i x^i$, write $P=(x+1)Q+R$ with constant $R$. Since $1+1=0$, <polynomial evaluation> at $x=1$ gives $P(1)=R$. This is the <polynomial remainder theorem> with $-1=1$ in the <field with two elements>. Hence
$$
\boxed{P(x)\bmod(x+1)=P(1)=p_0+p_1+\cdots+p_n.}
$$
The <polynomial remainder> is precisely the <parity bit> obtained by taking the <exclusive or> of all the <polynomial coefficients>.
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