Dividing polynomials

Polynomials can be divided, just like numbers. Division helps us factor polynomials and find their zeros. This page shows two methods: long division, which always works, and synthetic division, a shortcut for divisors of the form ${x-c}$.

The division algorithm

Start with numbers. When we divide ${17}$ by ${5}$, we get ${3}$ with remainder ${2}$. We can write this as

${17=5\cdot 3+2}$.

The remainder ${2}$ is smaller than the divisor ${5}$. Polynomials work the same way:

${P(x)=D(x)\cdot Q(x)+R(x)}$

Here ${P(x)}$ is the dividend (the polynomial we divide), ${D(x)}$ is the divisor, ${Q(x)}$ is the quotient, and ${R(x)}$ is the remainder. The remainder is either ${0}$, or its degree is less than the degree of ${D(x)}$.

Dividing both sides by ${D(x)}$ gives the same fact as a fraction:

$\dfrac{P(x)}{D(x)}=Q(x)+\dfrac{R(x)}{D(x)}$

Long division

Long division of polynomials follows the same steps as long division of numbers. Write both polynomials in descending powers. Then repeat these steps:

  1. Divide the leading term of what is left by the leading term of the divisor. This gives the next term of the quotient.
  2. Multiply the divisor by that term.
  3. Subtract the result, and bring down the next term.

Stop when the degree of what is left is less than the degree of the divisor. What is left is the remainder.

Example 1: Divide ${2x^3-7x^2+4x+5}$ by ${x-3}$.

Solution:

${2x^2}$${-x}$${+1}$
${x-3}$${2x^3}$${-7x^2}$${+4x}$${+5}$
${-}$${2x^3}$${-6x^2}$
${-x^2}$${+4x}$
${-}$${-x^2}$${+3x}$
${x}$${+5}$
${-}$${x}$${-3}$
${8}$

Here is each round of the work:

  • ${2x^3\div x=2x^2}$. Multiply: ${2x^2(x-3)=2x^3-6x^2}$. Subtract, and bring down ${+4x}$: this leaves ${-x^2+4x}$.
  • ${-x^2\div x=-x}$. Multiply: ${-x(x-3)=-x^2+3x}$. Subtract, and bring down ${+5}$: this leaves ${x+5}$.
  • ${x\div x=1}$. Multiply: ${1(x-3)=x-3}$. Subtract: this leaves ${8}$.

The degree of ${8}$ is ${0}$, less than the degree of ${x-3}$. So we stop. The quotient is ${2x^2-x+1}$, and the remainder is ${8}$:

$\begin{align*}&2x^3-7x^2+4x+5\\&=(x-3)(2x^2-x+1)+8\end{align*}$

Watch out: When you subtract, change the sign of every term in the row you subtract. For example, ${-7x^2-(-6x^2)=-x^2}$.

If a power of ${x}$ is missing in the dividend, write it with a coefficient of ${0}$. This keeps the columns lined up.

Example 2: Divide ${x^4+2x^3-x+3}$ by ${x^2+1}$.

Solution:

The dividend has no ${x^2}$ term, so write it as ${x^4+2x^3+0x^2-x+3}$. The divisor ${x^2+1}$ also has no ${x}$ term. So its multiples get a ${0x}$ or ${0x^3}$ term, too.

${x^2}$${+2x}$${-1}$
${x^2+1}$${x^4}$${+2x^3}$${+0x^2}$${-x}$${+3}$
${-}$${x^4}$${+0x^3}$${+x^2}$
${2x^3}$${-x^2}$${-x}$
${-}$${2x^3}$${+0x^2}$${+2x}$
${-x^2}$${-3x}$${+3}$
${-}$${-x^2}$${+0x}$${-1}$
${-3x}$${+4}$

The last line, ${-3x+4}$, has degree ${1}$. That is less than the degree ${2}$ of the divisor. So we stop. The quotient is ${x^2+2x-1}$, and the remainder is ${-3x+4}$:

$\begin{align*}&\dfrac{x^4+2x^3-x+3}{x^2+1}\\&=x^2+2x-1+\dfrac{-3x+4}{x^2+1}\end{align*}$

Check: Multiply the divisor by the quotient:

$\begin{align*}&(x^2+1)(x^2+2x-1)\\&=x^4+2x^3-x^2+x^2+2x-1\\&=x^4+2x^3+2x-1\end{align*}$

Then add the remainder:

$\begin{align*}&x^4+2x^3+2x-1+(-3x+4)\\&=x^4+2x^3-x+3\end{align*}$

This is the dividend, so the answer is right.

Synthetic division

When the divisor has the form ${x-c}$, there is a shortcut called synthetic division. It uses only the coefficients. Here is the method:

  1. Write ${c}$ on the left. Then write the coefficients of the dividend in a row. Use ${0}$ for every missing power.
  2. Bring the first coefficient down to the bottom row.
  3. Multiply that number by ${c}$, and write the product under the next coefficient. Add the two numbers, and write the sum in the bottom row.
  4. Repeat step 3 until you reach the end of the row.

The last number in the bottom row is the remainder. The other numbers are the coefficients of the quotient. The quotient has degree one less than the dividend.

Example 3: Use synthetic division to divide ${2x^3-7x^2+4x+5}$ by ${x-3}$.

Solution:

The divisor is ${x-3}$, so ${c=3}$. The coefficients are ${2}$, ${-7}$, ${4}$, and ${5}$.

${3}$${2}$${−7}$${4}$${5}$
${6}$${−3}$${3}$
${2}$${−1}$${1}$${8}$

Bring down ${2}$. Then ${2\cdot 3=6}$, and ${-7+6=-1}$. Next, ${-1\cdot 3=-3}$, and ${4+(-3)=1}$. Last, ${1\cdot 3=3}$, and ${5+3=8}$.

The bottom row ${2}$, ${-1}$, ${1}$ gives the quotient ${2x^2-x+1}$. The remainder is ${8}$. This is the same answer as in Example 1, with much less writing.

Watch out: Synthetic division uses ${c}$ from ${x-c}$. For the divisor ${x+2}$, write ${x+2=x-(-2)}$. So ${c=-2}$, not ${2}$.

Example 4: Divide ${x^4-5x^2+2x-8}$ by ${x+2}$.

Solution:

Here ${c=-2}$. There is no ${x^3}$ term, so the coefficients are ${1}$, ${0}$, ${-5}$, ${2}$, and ${-8}$.

${−2}$${1}$${0}$${−5}$${2}$${−8}$
${−2}$${4}$${2}$${−8}$
${1}$${−2}$${−1}$${4}$${−16}$

The dividend has degree ${4}$, so the quotient has degree ${3}$. The bottom row gives

${Q(x)=x^3-2x^2-x+4}$,

and the remainder is ${-16}$. So

$\begin{align*}&x^4-5x^2+2x-8\\&=(x+2)(x^3-2x^2-x+4)-16\end{align*}$

The remainder theorem

Look again at Example 4. Put ${x=-2}$ into the dividend:

$\begin{align*}&P(-2)\\&=16-20-4-8\\&=-16\end{align*}$

This is exactly the remainder. That is no accident:

Remainder theorem: If ${P(x)}$ is divided by ${x-c}$, the remainder is ${P(c)}$.

Why? The division algorithm says ${P(x)=(x-c)Q(x)+R}$. The remainder ${R}$ is a number, because its degree is less than ${1}$. Now put ${x=c}$. The factor ${c-c}$ is ${0}$, so ${P(c)=0\cdot Q(c)+R=R}$.

So synthetic division is also a quick way to find the value ${P(c)}$.

Example 5: Let ${P(x)=3x^5+5x^4-4x^3+7x+3}$. Use synthetic division to find ${P(-2)}$.

Solution:

Divide by ${x-(-2)}$, so ${c=-2}$. There is no ${x^2}$ term, so the coefficients are ${3}$, ${5}$, ${-4}$, ${0}$, ${7}$, and ${3}$.

${−2}$${3}$${5}$${−4}$${0}$${7}$${3}$
${−6}$${2}$${4}$${−8}$${2}$
${3}$${−1}$${−2}$${4}$${−1}$${5}$

The remainder is ${5}$. So ${P(-2)=5}$.

Check: Put ${x=-2}$ into ${P}$ directly. The powers are ${(-2)^5=-32}$, ${(-2)^4=16}$, and ${(-2)^3=-8}$. So the five terms of ${P(-2)}$ are ${3(-32)=-96}$, ${5(16)=80}$, ${-4(-8)=32}$, ${7(-2)=-14}$, and ${3}$.

Add them:

$\begin{align*}&P(-2)\\&=-96+80+32-14+3\\&=5\end{align*}$

The factor theorem

If the remainder is ${0}$, then ${P(x)=(x-c)Q(x)}$. So ${x-c}$ is a factor of ${P(x)}$. Together with the remainder theorem, this gives:

Factor theorem: ${x-c}$ is a factor of ${P(x)}$ exactly when ${P(c)=0}$.

So once we know one zero ${c}$, we can divide by ${x-c}$. The quotient has a lower degree, so it is easier to factor.

Example 6: Let ${P(x)=x^3-7x+6}$. Show that ${x-2}$ is a factor of ${P(x)}$. Then factor ${P(x)}$ completely, and find all its zeros.

Solution:

First, ${P(2)=8-14+6=0}$. So ${x-2}$ is a factor, by the factor theorem.

To find the other factor, divide by ${x-2}$. There is no ${x^2}$ term, so the coefficients are ${1}$, ${0}$, ${-7}$, and ${6}$:

${2}$${1}$${0}$${−7}$${6}$
${2}$${4}$${−6}$
${1}$${2}$${−3}$${0}$

The remainder is ${0}$, as expected. The quotient is ${x^2+2x-3}$, which factors as ${(x+3)(x-1)}$. So

${P(x)=(x-2)(x+3)(x-1)}$.

The zeros are ${2}$, ${-3}$, and ${1}$.

The factor theorem also works in reverse. Given the zeros, we can build the polynomial.

Example 7: Find a polynomial of degree ${3}$ with zeros ${-1}$, ${2}$, and ${3}$, and leading coefficient ${1}$.

Solution:

Each zero ${c}$ gives a factor ${x-c}$. So the factors are ${x+1}$, ${x-2}$, and ${x-3}$:

$\begin{align*}&P(x)\\&=(x+1)(x-2)(x-3)\\&=(x+1)(x^2-5x+6)\\&=x^3-5x^2+6x+x^2-5x+6\\&=x^3-4x^2+x+6\end{align*}$

Check: ${P(2)=8-16+2+6=0}$.

Summary

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