Dividing Polynomials

You already know how to divide. When you split 15 cookies among 3 people, you’re dividing. When you figure out how many $7 movie tickets you can buy with $50 (and realize you’ll have $1 left over), you’re dividing with a remainder. Dividing polynomials works the same way - you’re asking “how many times does this fit into that, and what’s left over?”

If regular long division felt mechanical in elementary school, polynomial long division might feel the same way at first. But here’s the good news: once you see the pattern, it becomes just as routine. And there’s even a shortcut called synthetic division that makes certain problems almost trivially fast.

Core Concepts

Why Divide Polynomials?

Before we dive into the mechanics, let’s talk about why you’d ever want to divide polynomials in the first place. There are several practical reasons:

Factoring: If you know that $(x - 2)$ is a factor of some polynomial $f(x)$, you can divide to find the other factor. This is like knowing that 3 is a factor of 15, then dividing to find the other factor is 5.

Finding zeros: Division helps you break down higher-degree polynomials into simpler pieces. If you can factor $x^3 - 6x^2 + 11x - 6$ into $(x - 1)(x - 2)(x - 3)$, you immediately know the zeros are 1, 2, and 3.

Simplifying rational expressions: When you have a fraction like $\frac{x^2 + 5x + 6}{x + 2}$, division can simplify it (in this case, to just $x + 3$).

Evaluating functions: The Remainder Theorem (which we’ll cover) gives you a surprisingly efficient way to calculate $f(c)$ for any polynomial function.

Polynomial Long Division

Polynomial long division follows the same algorithm as numerical long division. The key is to focus on the leading terms at each step.

Here’s the process:

1. Divide the leading term of the dividend by the leading term of the divisor
2. Multiply the entire divisor by that result
3. Subtract from the dividend
4. Bring down the next term
5. Repeat until the remainder has a lower degree than the divisor

Let’s see this in action. To divide $(x^2 + 5x + 6)$ by $(x + 2)$:

$$\require{enclose} \begin{array}{r} x + 3 \[-3pt] x + 2 \enclose{longdiv}{x^2 + 5x + 6} \[-3pt] \underline{-(x^2 + 2x)} \[-3pt] 3x + 6 \[-3pt] \underline{-(3x + 6)} \[-3pt] 0 \end{array}$$

Step by step:

• $x^2 \div x = x$ (first term of quotient)
• $x \cdot (x + 2) = x^2 + 2x$ (multiply)
• $(x^2 + 5x) - (x^2 + 2x) = 3x$ (subtract, bring down the 6)
• $3x \div x = 3$ (second term of quotient)
• $3 \cdot (x + 2) = 3x + 6$ (multiply)
• $(3x + 6) - (3x + 6) = 0$ (subtract)

The quotient is $x + 3$ with remainder 0. This means $(x^2 + 5x + 6) = (x + 2)(x + 3)$.

When there’s a remainder, we write the result as:

$$\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}$$

Or equivalently:

$$\frac{\text{Dividend}}{\text{Divisor}} = \text{Quotient} + \frac{\text{Remainder}}{\text{Divisor}}$$

Handling Missing Terms

What if your polynomial has “gaps” - missing powers of $x$? You need to include placeholders with coefficient 0.

For example, to divide $(x^3 - 8)$ by $(x - 2)$, you should write the dividend as $x^3 + 0x^2 + 0x - 8$. Those zeros keep your columns aligned, just like you’d write 305 instead of 35 when the tens place is empty.

Synthetic Division

When you’re dividing by a linear expression of the form $(x - c)$, there’s a much faster method called synthetic division. It uses only the coefficients and skips all the variable writing.

To use synthetic division for dividing by $(x - c)$:

1. Write $c$ (note: if dividing by $(x - 2)$, use $+2$; if dividing by $(x + 3)$, use $-3$)
2. Write the coefficients of the dividend (including 0s for missing terms)
3. Bring down the first coefficient
4. Multiply by $c$, add to the next coefficient
5. Repeat until done
6. The last number is the remainder; the others are coefficients of the quotient

Example: Divide $(x^3 - 8)$ by $(x - 2)$

Write it as $x^3 + 0x^2 + 0x - 8$, so coefficients are: 1, 0, 0, -8

$$\begin{array}{c|cccc} 2 & 1 & 0 & 0 & -8 \ & & 2 & 4 & 8 \ \hline & 1 & 2 & 4 & 0 \end{array}$$

Reading the result: quotient is $x^2 + 2x + 4$, remainder is 0.

This means $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$.

Why does synthetic division work? It’s actually doing the same thing as long division, just compressed. Each step is essentially dividing, multiplying, and subtracting - but since the divisor is always $(x - c)$, the pattern simplifies dramatically.

The Remainder Theorem

Here’s something beautiful: when you divide a polynomial $f(x)$ by $(x - c)$, the remainder equals $f(c)$.

The Remainder Theorem: If a polynomial $f(x)$ is divided by $(x - c)$, then the remainder is $f(c)$.

Think about why this works. When we divide, we get:

$$f(x) = (x - c) \cdot q(x) + r$$

where $q(x)$ is the quotient and $r$ is the remainder (a constant, since its degree must be less than the divisor’s degree of 1).

Now substitute $x = c$:

$$f(c) = (c - c) \cdot q(c) + r = 0 \cdot q(c) + r = r$$

So $f(c) = r$. The remainder is the function value.

This gives you a fast way to evaluate polynomials. Instead of substituting $c$ into $f(x)$ and computing all those powers, you can use synthetic division - especially nice for large values of $c$ or high-degree polynomials.

The Factor Theorem

The Factor Theorem is a direct consequence of the Remainder Theorem:

The Factor Theorem: $(x - c)$ is a factor of $f(x)$ if and only if $f(c) = 0$.

In other words, $c$ is a zero of the polynomial exactly when $(x - c)$ divides evenly into it.

This gives you two powerful tools:

• Testing potential factors: To check if $(x - 3)$ is a factor of $f(x)$, just compute $f(3)$. If you get 0, it’s a factor.
• Building from zeros: If you know $x = 2$ is a zero of $f(x)$, then $(x - 2)$ must be a factor.

Using Division to Factor Completely

Once you find one factor, division helps you find the rest. Here’s the strategy:

1. Find one zero (by testing values, graphing, or the Rational Root Theorem)
2. Divide the polynomial by $(x - c)$ where $c$ is your zero
3. Repeat with the quotient until you’ve reduced to linear or irreducible quadratic factors

This process works because each time you factor out $(x - c)$, the degree drops by one. A cubic becomes a quadratic, a quartic becomes a cubic, and so on.

Finding All Zeros Using Division

To find all zeros of a polynomial:

1. Find one zero $c$ (perhaps by inspection, graphing, or the Rational Root Theorem)
2. Divide by $(x - c)$ to get a lower-degree polynomial
3. Find zeros of the quotient
4. Repeat until you’ve found all zeros

For a degree-$n$ polynomial, you’ll have at most $n$ real zeros (counting multiplicity).

Notation and Terminology

Term	Meaning	Example
Dividend	Polynomial being divided	In $\frac{x^3 + 2x}{x - 1}$, dividend is $x^3 + 2x$
Divisor	What you’re dividing by	Divisor is $x - 1$
Quotient	Result of division	The polynomial you get
Remainder	What’s left over	Has degree less than divisor
Synthetic division	Shortcut for dividing by $(x - c)$	Uses only coefficients

Examples

Key Properties and Rules

Division Algorithm for Polynomials

For any polynomials $f(x)$ and $d(x)$ (where $d(x) \neq 0$), there exist unique polynomials $q(x)$ and $r(x)$ such that:

$$f(x) = d(x) \cdot q(x) + r(x)$$

where either $r(x) = 0$ or the degree of $r(x)$ is less than the degree of $d(x)$.

When to Use Each Method

Method	Use When	Advantages
Long division	Divisor is any polynomial	Always works
Synthetic division	Divisor is $(x - c)$	Much faster, fewer errors

Common Mistakes to Avoid

• Sign error in synthetic division: When dividing by $(x + 3)$, use $c = -3$, not $+3$
• Forgetting placeholders: Include 0 coefficients for missing terms
• Misreading the quotient: In synthetic division, the last number is the remainder - the others are the quotient coefficients
• Stopping too early: After dividing, check if the quotient can be factored further

Quick Reference: Synthetic Division Setup

Divisor	Value of $c$ to use
$(x - 2)$	$c = 2$
$(x + 5)$	$c = -5$
$(x - 7)$	$c = 7$
$(x + 1)$	$c = -1$

Real-World Applications

Simplifying Rational Expressions

When working with rational expressions (fractions with polynomials), division helps simplify. If the numerator is divisible by the denominator, you get a simpler polynomial. If not, you can still rewrite the expression in a more useful form.

For instance, $\frac{x^2 + 5x + 6}{x + 2} = x + 3$ (for $x \neq -2$).

Finding Exact Zeros of Polynomial Models

In physics, economics, and engineering, phenomena are often modeled by polynomials. Finding where these models equal zero (equilibrium points, break-even points, intersections) requires finding polynomial zeros. Division is the key tool for breaking down higher-degree polynomials once you’ve found one root.

Partial Fraction Decomposition

In calculus, you’ll need to break apart rational expressions into simpler fractions for integration. This process relies heavily on polynomial division and factoring. For example, integrating $\frac{x^3 + 2}{x^2 - 1}$ requires first doing polynomial division, then decomposing the remainder.

Algorithm Design

Computer algebra systems use polynomial division algorithms extensively. When you ask a calculator to factor a polynomial or find its zeros, it’s running division operations behind the scenes. Understanding the process helps you appreciate (and troubleshoot) what these tools are doing.

Self-Test Problems

Summary

• Polynomial division works like numerical long division: divide, multiply, subtract, bring down, repeat
• Synthetic division is a shortcut that only works when dividing by $(x - c)$. Use the value $c$ (watch the sign!) with just the coefficients
• The Remainder Theorem says that when you divide $f(x)$ by $(x - c)$, the remainder equals $f(c)$. This gives you a fast way to evaluate polynomials
• The Factor Theorem says $(x - c)$ is a factor of $f(x)$ if and only if $f(c) = 0$. Use this to test potential factors
• To factor completely, find one zero, divide it out, and repeat with the quotient
• Finding all zeros becomes systematic: test potential rational zeros, divide out the ones that work, factor what remains
• Always include placeholder zeros for missing terms when setting up division
• Check your work by multiplying: Divisor $\times$ Quotient $+$ Remainder should equal the original Dividend

Division transforms hard polynomial problems into easier ones. It’s the bridge between knowing one factor (or zero) and knowing them all. Once you’re comfortable with the mechanics, you’ll find it’s a reliable tool that opens up many other techniques in algebra and beyond.