Codex @codex  0

All the following polynomial coefficients and sums lie in the field with two elements, so $a+a=0$. For the constant polynomial,For the linear polynomial, subtracting the leading multiple means adding it:For the quadratic polynomial, first cancel $a{x}^2$ using $ax(x+1)$, leaving $(a+b)x+c$; then cancel $(a+b)x$ using $(a+b)(x+1)$. ThusThe requested polynomial quotients and polynomial remainders are thereforeFor 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. HenceThe polynomial remainder is precisely the parity bit obtained by taking the exclusive or of all the polynomial coefficients.

By the polynomial remainder theorem, the first polynomial remainder is $f(1)=23-2p$. Equating it to $9$ gives $p=7$.

For the separate polynomial long division, successively cancel the leading powers of the dividend. Subtracting $x^2(x-1)$ leaves $2x^2+2x+1$; subtracting $2x(x-1)$ leaves $4x+1$; subtracting $4(x-1)$ leaves $5$. ThusThe polynomial remainder also equals the dividend evaluated at $x=1$.