Polynomial Division

State the division identity

For polynomials $P$ and $D$ with real coefficients and $D\ne0$, there are unique polynomials $Q$ and $R$ such that

where either $R=0$ or $\deg R<\deg D$. This is the polynomial division algorithm. The restriction concerns the degree of a nonzero remainder, not the numerical size of its coefficients. A constant remainder of 100 is acceptable when the divisor is linear.

When $D(x)\ne0$, divide the identity by $D(x)$:

These two statements have different domains. The polynomial identity holds for every real $x$. The quotient expression is undefined at zeros of the divisor, even if the remainder is zero. Keep that distinction when moving from polynomial algebra to a rational function.

If the dividend has lower degree than the divisor, take $Q=0$ and $R=P$. A nonzero constant divisor gives a zero remainder. The zero polynomial is allowed as a dividend or remainder; it does not need a degree assigned here. OpenStax’s division treatment provides the standard algorithm and notation.

Divide by cancelling the leading term

Divide

by $D(x)=x+2$. Write terms in descending powers. Each step removes the current highest power.

First term of the quotient

Divide $2x^3$ by $x$ to obtain $2x^2$. Multiply the divisor by this quotient term, then subtract the entire product:

The cubic term has disappeared. The next quotient term must remove the remaining quadratic term.

Continue until the remainder has lower degree

Divide $-x^2$ by $x$ to obtain $-x$. The subtraction is

Finally, divide $-3x$ by $x$ to obtain $-3$:

A constant has lower degree than a linear divisor, so the procedure stops. Thus

Reconstruct the original dividend:

This expansion checks every coefficient, rather than merely checking one selected input. As a rational expression, the result is $2x^2-x-3+13/(x+2)$, with $x\ne-2$.

Use synthetic division as a coefficient algorithm

For a divisor $x-c$, synthetic division records the same calculation using coefficients. In the previous example, $x+2=x-(-2)$, so use $c=-2$, not $2$.

Bring down 2. Multiply by −2 and add to 3, giving −1. Multiply −1 by −2 and add to −5, giving −3. Multiply −3 by −2 and add to 7, giving 13. The first three running totals form the quotient coefficients; the last is the remainder.

The shorter notation does not remove the need to align powers. For $x^3+1$, the coefficients are $1,0,0,1$. Omitting the two zeros would change the polynomial being divided.

Handle a non-monic linear divisor correctly

Consider

Long division gives the quotient $2x^2+x-3$ and remainder zero. One check is

A synthetic calculation at $c=1/2$ divides by $x-1/2$, not by $2x-1$. It gives the quotient $4x^2+2x-6$. Since $2x-1=2(x-1/2)$, divide that quotient by 2 to obtain the quotient for the original divisor.

The remainder is unchanged by this scaling. More generally, if

then

This explains both adjustments. Substituting $c$ finds the correct remainder for either linear divisor, but their quotient polynomials differ by a factor of $a$.

Allow a polynomial remainder for a higher-degree divisor

Divide $x^4+2x^3+3x^2+4x+5$ by $x^2+1$. Successive leading-term cancellations use $x^2$, $2x$ and $2$:

The result is

Stop at the linear remainder. Requiring a constant would impose a condition that belongs only to linear divisors. Here $x^2+1$ is nonzero for every real input, so the associated real quotient has no excluded input.

Understand why the answer is unique

Repeated leading-term cancellation lowers the degree at each step, so it must stop. This establishes the procedure’s existence over the real numbers, where dividing by a nonzero leading coefficient is possible.

For uniqueness, suppose two valid answers satisfy $P=DQ_1+R_1=DQ_2+R_2$. Subtract:

If $Q_1-Q_2$ were nonzero, the left side would have degree at least $\deg D$. The right side is either zero or has degree below $\deg D$. They cannot be the same nonzero polynomial. Therefore the quotients agree, and then the remainders agree.

That argument also explains the stopping rule. Without the degree restriction, many product-plus-remainder descriptions would be possible.

Check algebra and interpretation separately

Check the identity by expanding $DQ+R$. Check the stopping rule by comparing degrees. Finally, check the domain when the answer is written as a quotient. These tests answer different questions.

A zero remainder establishes a factor. It does not assign a value to $P/D$ at a zero of $D$. An omitted zero coefficient changes the synthetic calculation. A sign mistake while subtracting a product usually contaminates all later coefficients, so keep parentheses around that product until the subtraction is complete.

Related topics

Use the Remainder Theorem when only a remainder is needed. The Factor Theorem turns zero remainders into factorisations, and Asymptotes uses the quotient to describe end behaviour.