How to know what is the degree of the remainder

Question

The example question is

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

The answer did something like

$$8x^4+3x-1=(2x^2+1)(Ax^2+Bx+C)+(Dx+E)$$

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

Answer 1

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

Answer 2

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.

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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 $$\begin{matrix} 4x^2 - 2\\ \qquad\qquad\qquad 2x^2+1\ \overline{ )\ 8x^4 \; +0x^3 \; +0x^2 \; + 3x - 1 }\\ \qquad\qquad\qquad \underline{ 8x^4 \; +0x^3 \;+ \;\;4x^2}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\; -4x^2\; + 3x - 1 \\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\; \underline{4x^2\; + 0x - 2}\\ \qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad\qquad\qquad\qquad\qquad 3x + 1\\ \end{matrix}\tag{8}$$

Answer 3

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 $$ Q(X^{-1})=T_0^q\Big(f(X^{-1})X^q\Big)X^{-q}. $$