Synthetic Division

Learn a faster method for dividing polynomials by linear factors using synthetic division.

Advanced25 minLesson

Definition

Synthetic division is a shortcut method for dividing a polynomial by a linear binomial of the form

(x - c)

Instead of using long division, we work only with the coefficients and perform simple arithmetic operations.

When to use synthetic division:

The divisor must be linear: $(x - c)$ or $(x + c)$
The coefficient of $x$ in the divisor must be $1$

The setup: To divide

P(x)

(x - c)

1.Write the value of $c$ (the root) on the left
2.Write the coefficients of $P(x)$ in order
3.Include $0$ for any missing terms

\begin{array}{c|cccc} c & a_n & a_{n-1} & \cdots & a_0 \\ \end{array}

Try it now

When dividing by

(x - 5)

, what value of

c

do you use in synthetic division?

Worked Examples

Divide $(x^3 - 6x^2 + 11x - 6)$ by $(x - 2)$

Identify the divisor value

$(x - 2)$ means $c = 2$

Write coefficients in order

$x^3 - 6x^2 + 11x - 6$ has coefficients: $1, -6, 11, -6$ → $1 \quad -6 \quad 11 \quad -6$

Set up the synthetic division

\begin{array}{c|cccc} 2 & 1 & -6 & 11 & -6 \\ & & & & \\ \hline & & & & \end{array}

→ Ready to calculate

Bring down the first coefficient

\begin{array}{c|cccc} 2 & 1 & -6 & 11 & -6 \\ & & & & \\ \hline & 1 & & & \end{array}

→ First coefficient:

1

Multiply and add for each column

$2 \times 1 = 2$ , then $-6 + 2 = -4$ $2 \times (-4) = -8$ , then $11 + (-8) = 3$ $2 \times 3 = 6$ , then $-6 + 6 = 0$ →

\begin{array}{c|cccc} 2 & 1 & -6 & 11 & -6 \\ & & 2 & -8 & 6 \\ \hline & 1 & -4 & 3 & 0 \end{array}

Write the quotient and remainder

Bottom row: $1, -4, 3$ are quotient coefficients; $0$ is remainder → Quotient: $x^2 - 4x + 3$ , Remainder: $0$

Answer: $(x^3 - 6x^2 + 11x - 6) \div (x - 2) = x^2 - 4x + 3$ Since the remainder is $0$ , $(x - 2)$ is a factor of the polynomial.

Common Mistakes

Forgetting to include zeros for missing terms

Why it's wrong: Every power of $x$ from highest to constant must have a coefficient. Missing terms have coefficient $0$ .

Correct: For $x^4 - 16$ , write: $1, 0, 0, 0, -16$ (zeros for $x^3$ , $x^2$ , and $x$ )

Using the wrong sign for c when dividing by $(x + c)$

Why it's wrong: Synthetic division uses the root form $(x - c)$ . When dividing by $(x + 2)$ , you need $c = -2$ .

Correct: $(x + 2) = (x - (-2))$ , so use $c = -2$ in the synthetic division setup

Adding when you should multiply, or vice versa

Why it's wrong: The pattern is: multiply by $c$ , then add to the next coefficient.

Correct: Always: bring down first coefficient, then repeat (multiply by $c$ , add to next coefficient)

Writing the quotient with the wrong degree

Why it's wrong: The quotient has one degree less than the original polynomial.

Correct: If dividing a cubic ( $x^3$ ) by linear, the quotient is quadratic ( $x^2$ )

Watch a video explanation

The same topic explained by another teacher, if a video helps you more.

Synthetic Division of Polynomials: Step-by-Step Examples
by Understand The Math
Watch on YouTube

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Practice Problems

18 problems

Problem 1 of 18

Easy

When dividing by $(x - 5)$ , what value of $c$ do you use in synthetic division?

Why It Matters

Synthetic division is a powerful tool that saves time and reduces errors:

Speed: Much faster than polynomial long division
Simplicity: Only uses basic arithmetic (addition and multiplication)
Factoring: Helps find roots of polynomials quickly
Remainder Theorem: If you divide $P(x)$ by $(x - c)$ , the remainder equals $P(c)$

Real applications:

Engineering: Analyzing transfer functions in control systems
Computer graphics: Polynomial curve calculations
Finance: Modeling growth patterns with polynomial functions

Real World Applications

Engineering: Transfer Functions

Engineers use polynomial division when analyzing control systems. The transfer function of a system is often a ratio of polynomials.

Example:

If a system has transfer function $\frac{x^3 - 6x^2 + 11x - 6}{x - 2}$ , synthetic division simplifies it to $x^2 - 4x + 3$ .

1Try It Yourself

A filter circuit has polynomial response $P(x) = x^3 + 2x^2 - 5x - 6$ . You need to factor it by testing if $x = 2$ is a root.

Use synthetic division to test if $(x - 2)$ is a factor.

Step 1: Write the mathematical expression

Set up: coefficients are $1, 2, -5, -6$ with $c = 2$

Computer Science: Algorithm Optimization

Evaluating polynomials efficiently uses ideas from synthetic division. Horner's method is essentially synthetic division for function evaluation.

Example:

To find $P(5)$ for $P(x) = 2x^3 - x^2 + 3x - 7$ , use synthetic division with $c = 5$ . The remainder equals $P(5) = 213$ .

2Try It Yourself

You need to evaluate $P(x) = x^3 - 4x + 2$ at $x = 3$ for a graphics calculation.

Use synthetic division to find $P(3)$ .

Step 1: Write the mathematical expression

Coefficients: $1, 0, -4, 2$ (remember the missing $x^2$ term!)

Key Takeaways

1Synthetic division is a shortcut for dividing polynomials by $(x - c)$
2Use only the coefficients, including $0$ for missing terms
3Pattern: bring down, multiply by $c$ , add, repeat
4The last number is the remainder; other numbers are quotient coefficients
5For $(x + c)$ , use $c = -c$ in the setup
6Remainder Theorem: the remainder equals $P(c)$

Frequently Asked Questions

No, synthetic division only works when dividing by a linear binomial

(x - c)

where the coefficient of

x

1

. For other divisors like

(2x - 3)

(x^2 + 1)

, use polynomial long division.

Glossary

Synthetic division: A shortcut method for dividing a polynomial by a linear binomial $(x - c)$ using only coefficients
Divisor: The polynomial you are dividing by; in synthetic division, it must be of the form $(x - c)$
Quotient: The result of the division (excluding the remainder)
Remainder: The amount left over after division; equals $P(c)$ by the Remainder Theorem
Remainder Theorem: States that when $P(x)$ is divided by $(x - c)$ , the remainder equals $P(c)$
Factor Theorem: States that $(x - c)$ is a factor of $P(x)$ if and only if $P(c) = 0$

Formula Card

Synthetic Division Setup

\begin{array}{c|ccc} c & a_n & a_{n-1} & \cdots & a_0 \end{array}

Write $c$ on left, coefficients in descending order on right

The Algorithm

Bring down

\to

Multiply by

c

\to

Add

\to

Repeat

Continue until all coefficients are processed

Result Interpretation

P(x) = (x-c) \cdot Q(x) + R

Last number is remainder $R$; others form quotient $Q(x)$

Remainder Theorem

P(c) = R

The remainder equals the polynomial evaluated at $c$