*Polynomial division* is the process whereby a polynomial, $p(x)$, is divided by a polynomial, $d(x)$, usually of degree less than or equal to that of $p(x)$. The result is of the form \[\frac{p(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)}\]

where $q(x)$ is the *quotient polynomial* and $r(x)$ the *remainder polynomial* of degree less than the degree of $d(x)$.

If $r(x)=0$ then $d(x)$ is said to divide $p(x)$ exactly.

There are two common methods of polynomial division: *long division* and *factorisation*.

The process is very like long division of numbers, except with algebra. This method is best explained using an example.

Divide $x^3-3x^2+x+1$ by $x-1$.

Begin by writing the question out in long division form:

\begin{align} x-1\overline{\big)x^3\!-3x^2\!+x\!+\!1} \end{align}

Starting with the first term in the polynomial, $x^3$, calculate what $x-1$ needs to be multiplied by so that when we subtract it from the polynomial, it will eliminate the $x^3$ term. In this case it is $x^2$, as $x^2 \times (x-1) = x^3 -x^2$, so we **subtract** this from the polynomial under the division sign. We write the $x^2$ on top of the division line.

\begin{align} &\;\: x^2\\ x-1&\overline{\big)x^3\!-3x^2\!+x\!+\!1}\\ &\;\: x^3-x^2\\ \end{align}

\begin{align} &\;\:x^2\\ x-1&\overline{\big)x^3\!-3x^2\!+x\!+\!1}\\ &\;\:x^3-x^2\\ &\;\:\overline{0 -2x^2+x+1} \end{align}

We now have a new polynomial $-2x^2+x+1$ with which to repeat the process.

This time we are looking to eliminate $-2x^2$, so we multiply $x-1$ by $-2x$ to get $-2x^2 +2x$ which we can then subtract to get rid of the $x^2$ term.

\begin{align} &\;\:x^2-2x\\ x-1&\overline{\big)x^3\!-3x^2\!+x\!+\!1}\\ &\;\:x^3-x^2\\ &\;\:\overline{0 -2x^2+x+1}\\ &\quad\;\; -2x^2 +2x \end{align}

\begin{align} &\;\:x^2-2x\\ x-1&\overline{\big)x^3\!-3x^2\!+x\!+\!1}\\ &\;\:x^3-x^2\\ &\;\:\overline{0 -2x^2+x+1}\\ &\quad\; -2x^2 +2x\\ &\qquad \overline{\; 0\:-\:x\:+\:1} \end{align}

Finally, we look to eliminate $-x$ so we multiply our divisor by $-1$ to get $-x+1$, and subtract.

\begin{align} &\;\:x^2-2x-1\\ x-1&\overline{\big)x^3\!-3x^2\!+x\!+\!1}\\ &\;\:x^3-x^2\\ &\;\:\overline{0 -2x^2+x+1}\\ &\quad\; -2x^2 +2x\\ &\qquad \overline{\; 0\:-\:x\:+\:1}\\ &\qquad \quad \; -\:x \;+\;1\\ &\qquad \qquad \qquad \;\overline{\;0} \end{align}

This happens to have eliminated the constant as well and we are left with zero. This means that $x-1$ is a *factor* of the cubic. We have shown \[\frac{x^3-3x^2+x+1}{x-1}=x^2-2x-1\]

Divide $3x^3-5x^2+10x-3$ by $3x+1$.

Begin by writing the question out in long division form.

\begin{align} 3x+1\overline{\big)3x^3\!-5x^2\!+10x\!-\!3} \end{align}

Starting with the first term in the polynomial, $3x^3$, calculate what $3x+1$ needs to be multiplied by so that when we subtract it from the polynomial, it will eliminate the $x^3$ term. In this case it is $x^2$, as $x^2 \times (3x+1) = 3x^3 +x^2$, so we **subtract** this from the polynomial under the division sign. We write the $x^2$ on top of the division line.

\begin{align} &\;\: x^2\\ 3x+1&\overline{\big)3x^3\!-5x^2\!+10x\!-\!3}\\ &\;\: 3x^3+x^2\\ \end{align}

\begin{align} &\;\: x^2\\ 3x+1&\overline{\big)3x^3\!-5x^2\!+10x\!-\!3}\\ &\;\: 3x^3+x^2\\ &\;\:\overline{0 -6x^2+10x-3} \end{align}

We now have a new polynomial $-6x^2+10x-3$ with which to repeat the process.

This time we are looking to eliminate $-6x^2$, so we multiply $3x+1$ by $-2x$ to get $-6x^2 -2x$ which we can then subtract to get rid of the $x^2$ term.

\begin{align} &\;\: x^2\,-\,2x\\ 3x+1&\overline{\big)3x^3\!-5x^2\!+10x\!-\!3}\\ &\;\: 3x^3+x^2\\ &\;\:\overline{0 -6x^2+10x-3}\\ &\quad\;\; -6x^2 -2x \end{align}

\begin{align} &\;\: x^2\,-\,2x\\ 3x+1&\overline{\big)3x^3\!-5x^2\!+10x\!-\!3}\\ &\;\: 3x^3+x^2\\ &\;\:\overline{0 -6x^2+10x-3}\\ &\quad\;\; -6x^2 -2x\\ &\qquad \overline{\; 0\:+ \: 12x\:-\:3} \end{align}

Now, we look to eliminate $12x$ so we multiply our divisor by $4$ to get $12x+4$, and subtract.

\begin{align} &\;\: x^2\,-\,2x\,+\,4\\ 3x+1&\overline{\big)3x^3\!-5x^2\!+10x\!-\!3}\\ &\;\: 3x^3+x^2\\ &\;\:\overline{0 -6x^2+10x-3}\\ &\quad\;\; -6x^2 -2x\\ &\qquad \overline{\; 0\:+\: 12x\:-\:3}\\ &\qquad \qquad \;\; 12x \,+\,4\\ &\qquad \qquad \qquad \; \;\overline{\;-7} \end{align}

We cannot divide this any more and are left with a remainder of $-7$. So \[\frac{3x^3-5x^2+10x-3}{3x+1}=x^2-2x+4 - \frac{7}{3x+1}\]

Prof. Robin Johnson uses the long division method to divide $2x^3+x^2-4x+4$ by $x+2$.

In this example, Prof Johnson finds all the roots of a cubic equation, with one of the roots already known, by first finding a factor of the cubic given by the known root and then dividing by that factor to obtain a quadratic and then find its roots to find the remaining two roots of the cubic.

The factorisation method involves *forcing* each term in $p(x)$ to have a factor $d(x)$.

Consider the case when $d(x)$ is a polynomial of degree one, i.e. of the form $x+b$. For each term $a_nx^n$ in $p(x)$, replace it with $a_nx^{n-1}(x+b) - a_nx^{n-1}b$. This gives us one term with $x+b$ as a factor, and a second term of lower degree which can be combined with the other $x^{n-1}$ term in the original expression. After applying this method to each non-constant term, we end up with an expression whose terms all have a factor of $d(x)$, apart from the constant term (if there is one).

Divide $x^2-9x-10$ by $x+1$.

First, we want to take a factor of $x$ out of the $x^2$ term and replace it with $(x+1)$. $x(x+1) = x^2+x$, so we need to subtract an $x$ term in order to have the same quantity as we started with. We now have

\begin{align} x^2 -9x -10 &= x(x+1) - x -9x -10 \\ &= x(x+1) - 10x - 10 \end{align}

Now we consider the $-10x$ term. As before, we take out a factor of $x$ and replace it with $(x+1)$. $-10(x+1) = -10x - 10$, so we need to add $10$ to preserve equality. We now have

\begin{align} x(x+1) -10x -10 &= x(x+1) -10(x+1) +10 - 10 \\ &= x(x+1) -10(x+1) \end{align}

Now every term in the expression has a factor of $(x+1)$ so it's straightforward to divide by $x+1$.

\[\frac{x^2-9x-10}{x+1} = \frac{x(x+1)-10(x+1)}{x+1} = x-10.\]

**Note:** There is no remainder, so we have shown that $x+1$ is a factor of $x^2-9x-10$.

Divide $x^3-3x^2+5x-4$ by $x-2$.

First, consider the term $x^3$. Take out a factor of $x$ and replace it with $x-2$. $x^2(x-2) = x^3-2x^2$, so we must add $2x^2$ to preserve equality.

\begin{align} x^3 - 3x^2 + 5x - 4 &= x^2(x-2) +2x^2 -3x^2 +5x - 4 \\ &= x^2(x-2) - x^2 + 5x - 4 \end{align}

Now consider the $-x^2$ term. This is replaced with $-x(x-2) - 2x$.

\begin{align} x^2(x-2) -x^2 +5x - 4 &= x^2(x-2) -x(x-2) - 2x + 5x - 4 \\ &= x^2(x-2) - x(x-2) + 3x - 4 \end{align}

Finally, consider the $3x$ term. This is replaced with $3(x-2) + 6$.

\begin{align} x^2(x-2) - x(x-2) + 3x - 4 &= x^2(x-2) - x(x-2) +3(x-2) + 6 - 4 \\ &= x^2(x-2) - x(x-2) +3(x-2) + 2 \end{align}

Now as many terms as possible have a factor of $x-2$, so it's straightforward to perform the division:

\begin{align} \frac{x^3 - 3x^2 + 5x - 4}{x-2} &= \frac{x^2(x-2) -x(x-2) +3(x-2) +2}{x-2} \\ &= x^2 - x +3 + \frac{2}{x-2} \end{align}

Prof. Robin Johnson uses the factorisation method to divide $x^3-3x^2+x+1$ by $x-1$, and $2x^3 +5x^2-3x+2$ by $x+2$.

In this example, Prof Johnson finds all the roots of the cubic equation $x^3-2x^2-21x-26=0$ with one of the roots already known, by first finding a factor $x+2$ of the cubic $x^3-2x^2-21x-26$ given by the known root $x=-2$ and then dividing by that factor to obtain a quadratic. Finally, he solves the quadratic to give the remaining two roots of the cubic.

Test yourself: Numbas test on polynomial division

- Polynomial division workbook at
**math**centre.