The example question is

Find the remainder when $8x^4+3x-1$ is divided by $2x^2+1$

The answer did something like


Where $(Ax^2+Bx+C)$ is the Quotient and $(Dx+E)$ the remainder. I believe the degree of Quotient is derived from degree of $8x^4+3x-1$ - degree of divisor. But for remainder? Would it not be

  • 5
    $\begingroup$ The degree of the remainder will be at most one less than the degree of the divisor. If you assume otherwise, then you could just multiply the divisor by the necessary factor to knock out the "remainder"s term of largest degree and then you're left with a new remainder of lesser degree, contradicting the hypothesis you actually finished the division. $\endgroup$
    – anon
    Sep 18, 2011 at 10:36
  • $\begingroup$ Also, there's a straightforward method called polynomial long division which allows one to find the quotient and remainder without tediously (and potentially aimlessly) solving for any introduced variables but rather simply following a fixed set of arithmetical rules. $\endgroup$
    – anon
    Sep 18, 2011 at 10:40
  • $\begingroup$ This question has nothing to do with division algebras. Replaced that with a more appropriate tag. $\endgroup$ Sep 18, 2011 at 14:25
  • $\begingroup$ @anon: I think the "tediously (and potentially aimlessly) solving for any introduced variables" technique is worthy knowing: it is commonly used to find partial fraction expansions. $\endgroup$ Sep 18, 2011 at 16:39
  • 1
    $\begingroup$ @Americo: That's a good point. I suppose it ultimately becomes a linear algebra problem in either case, which is straightforward. I just remember when I first started out using "introduce new variables and solve for them all" all over the place in math and it didn't go very efficiently for me. Guess I'm projecting. :) $\endgroup$
    – anon
    Sep 18, 2011 at 19:33

3 Answers 3


Polynomial division allows for a polynomial to be written in a divisor–quotient form:

$\frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$, where degree(D) < degree(P) and degree(R) < degree(D)

This rearrangement is known as the division transformation.

In this particular case $R(x)=3x-3$ so degree(R)=1 < degree(D)=2

  • $\begingroup$ If $\text{degree}(D) > \text{degree} (P)$, then $Q(x)=0$ and $P(x)=R(x)$. $\endgroup$ Sep 18, 2011 at 16:17

EDIT to add these short answers.

I believe the degree of Quotient is derived from degree of $8x^4+3x-1$ - degree of divisor.

That's right.

But for remainder?

The degree of the remainder is less than the degree of the divisor, by definition of polynomial division.

  1. Let me start with this specific case. To find the remainder ($Dx+E$) we can expand the RHS of the identity shown in the question $$8x^4+3x-1=(2x^2+1)(Ax^2+Bx+C)+(Dx+E)\tag{1}$$ and collect the terms of the same degree. We get $$8x^4+3x-1=2Ax^4+2Bx^3+(A+2C)x^2+(B+D)x+C+E.\tag{2}$$ The polynomial of the LHS is equivalent to to polynomial of the RHS if and only if the coefficients of the terms of the same degree are equal. Therefore we must have the following system of $5$ equations $$2A=8,\ 2B=0, \ A+2C=0, \ B+D=3, \ C+E=-1, $$ whose solution is $$A=4, \ B=0, \ C=-2, \ D=3, \ E=1. \ $$ Hence we obtain $$8x^4+3x-1=(2x^2+1)(4x^2-2)+(3x+1),\tag{3}$$ where $3x+1$ is the remainder. The degree of $8x^4+3x+1$ is $4$ and the degree of $2x^2+1$ is $2$. The degree of the quotient $(4x^2-2)$ is $2=4-2$. The degree of the remainder is $1<2$, which means that it is less than the degree of the divisor $2x^2+1$. Since $2x^2+1\ne 0$, the algebraic identity $(3)$ is equivalent to $$\frac{8x^4+3x-1}{2x^2+1}=4x^2-2+\frac{3x+1}{2x^2+1}.\tag{4}$$
  2. Now the general case. By definition of polynomial division, given polynomials $A(x),B(x)$, where the degree of $B(x)$ is greater than $0$, it is always possible to find a polynomial $Q(x)$, called quotient, such that the diference $$R(x)=A(x)-B(x)Q(x)\tag{5}$$ is a polynomial whose degree is less than the degree of $B(x)$. This polynomial $R(x)$, called remainder, is unique. The polynomial $A(x)$ is called the dividend and $B(x)$ the divisor. Let $m$ be the degree of $A(x)$, $n$ the degree of $B(x)$ and $q$ the degree of $Q(x)$. If $m<n$, $Q(x)=0$ and $R(x)=A(x)$. If $m\ge n$, then $q=m-n$. (Note: if $n=0$, then $R(x)=0$.) The identity $(5)$ is equivalent to $$A(x)=B(x)Q(x)+R(x)\tag{6}$$ and for $B(x)\ne 0$ to $$\frac{A(x)}{B(x)}=Q(x)+\frac{R(x)}{B(x)}.\tag{7}$$
  3. Concerning the computation of the quotient and the remainder, in addition to the method detailed above, we can use the polynomial long division or the synthetic division. The long division technique applied to the present case, results in $$\begin{matrix} 4x^2 - 2\\ \qquad\qquad\qquad 2x^2+1\ \overline{ )\ 8x^4 \; +0x^3 \; +0x^2 \; + 3x - 1 }\\ \qquad\qquad\qquad \underline{ 8x^4 \; +0x^3 \;+ \;\;4x^2}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\; -4x^2\; + 3x - 1 \\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\; \underline{4x^2\; + 0x - 2}\\ \qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad\qquad\qquad\qquad\qquad 3x + 1\\ \end{matrix}\tag{8}$$

There is a simple closed formula for the remainder $R$ and the quotient $Q$ of the euclidean division of a polynomial $P$ by a nonzero polynomial $D$. Here $P,D,Q,R$ are in $\mathbb C[X]$.

For any complex number $a$, any nonnegative integer $k$, and any rational fraction $f(X)\in\mathbb C(X)$ defined at $a$, let $$T_a^k(f(X))$$ be the degree at most $k$ Taylor approximation of $f(X)$ at $X=a$.

We may assume $$ D(X)=\big(X-a_1\big)^{m_1}\cdots\big(X-a_r\big)^{m_r}, $$ where the $a_j$ are distinct and the $m_j$ positive. Then we have $$ R(X)=\sum_{j=1}^r\ T_{a_j}^{m_j-1}\left(P(X)\ \frac{(X-a_j)^{m_j}}{D(X)}\right) \frac{D(X)}{(X-a_j)^{m_j}}\quad. $$ If $m_j=1$ for all $j$, we get Lagrange's Interpolation Formula $$ R(X)=\sum_{j=1}^r\ P(a_j)\ \prod_{k\not=j}\ \frac{X-a_k}{a_j-a_k}\quad. $$ If $\deg P < \deg D$, then $Q=0$. Otherwise, putting $q:=\deg P-\deg D$ and $f:=P/D$, we have $$ Q(X^{-1})=T_0^q\Big(f(X^{-1})X^q\Big)X^{-q}. $$


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