Does there exist an explicit formula for the result of polynomial division with non linear factors?

Question

The remainder theorem states that, if a polynomial $P(x)$ is divided by a linear factor $x-b$ , then the remainder of the division is $ P(b)$. I am looking for a more general result of what would be the outcome if the dividing factor was some general factor instead of a linear one.. i.e:

$$ P(x) = R(x) + Q(x) J(x)$$

Where, $ R(x)$ is the remainder, $J(x)$ is the dividing polynomial which has degree less then $P(x)$ and $Q(x)$ is the divisor. I have previously derived this result for factor of the form See here:

$$J(x) = (x-a)^n$$

but suppose I had some irreducible polynomial of the form( as an example):

$$J= x^2 + x +1$$

Then what would be the remainder?

Answer 1

The remainder will be $0$ or a polynomial with degree less than $x^2+x+1$. For example, if we divide $x^4$ by $x^2+x+1$: $$ \require{enclose} \begin{array}{r} \color{#C00}{1x^2}\color{#090}{-1x}\color{#00F}{+0}\\[-4pt] x^2+x+1\enclose{longdiv}{x^4+0x^3+0x^2+0x+0}\\[-4pt] \underline{\color{#C00}{x^4+1x^3+1x^2}}\phantom{+0x+0\ \ }\\[-4pt] -1x^3-1x^2\phantom{+0x+0\ \ }\\[-4pt] \underline{\color{#090}{-1x^3-1x^2-1x}}\phantom{+0\ \ }\\[-4pt] 1x+0\\[-4pt] \underline{\color{#00F}{0x^2+0x+0}}\\[-4pt] 1x+0\\[-4pt] \end{array} $$ Thus, $$ x^4=\left(x^2+x+1\right)\overbrace{\left(x^2-x\right)}^Q+\overset{\substack{R\\\downarrow\\{\phantom{A}}}}{x} $$

Answer 2

Consider $(x^3 + x^2 - x + 2):( x^2 + 1)$

The remainder has the form $R(x)=ax+b$. Divisor is $(x+i)(x-i)$

$x^3 + x^2 - x + 2=Q(x)(x+i)(x-i)+ax+b$

Plug $x=i$

$i^3+i^2-i+2=Q(i)(i+i)(i-i)+ai+b \to ai+b=1-2 i$

and $x=-i\to -ai+b=1+2 i$

we've got the system $$ \begin{cases} ai+b=1-2 i\\ -ai+b=1+2 i\\ \end{cases} $$ $b=1,a=-2$

The remainder is $R(x)=-2x+1$

No general rule AFAIK