Why is Synthetic Division faster than Polynomial Long Division?

Synthetic Division compresses the algorithm into a single compact table and works with only the coefficients, not the variable x. You avoid writing the intermediate quotient and remainder at each step, reducing both the writing and the chance of arithmetic errors.

Can I use Synthetic Division if the divisor has a leading coefficient other than 1?

No, Synthetic Division only works for divisors of the form (x - c). If your divisor is (2x - 1) or (3x + 2), factor out the leading coefficient to rewrite it as 2(x - \frac{1}{2}) or 3(x - \frac{2}{3}), then either adjust the final quotient or use Polynomial Long Division instead.

What does the last number in the bottom row of the synthetic division table represent?

By the Remainder Theorem, that last number is the remainder, and it also equals P(c)—the value of the polynomial at x = c. If the remainder is zero, then (x - c) is a factor of P(x).

What if my polynomial is missing terms, like x^3 - 2?

Write the coefficients including zeros for missing powers: x^3 + 0x^2 + 0x - 2 has coefficients 1, 0, 0, -2. If you skip the zeros, the products and additions will not line up correctly with the original polynomial's terms.

Synthetic Division

A fast method for dividing a polynomial by a binomial of the form $(x - c)$ , giving quotient and remainder efficiently without long division.

P(x) \div (x - c)

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Type the problem. The solver will use Synthetic Division where Synthetic Division is the right tool, and tell you when it is not.

What each symbol means

What Synthetic Division takes

Symbol	Meaning
$P$	The polynomial being divided; $P(x)$ represents the complete polynomial expression you start with. If you misidentify which polynomial is $P$ , the entire division will be wrong.
$x$	The variable in the polynomial; the letter representing the unknown quantity in the expression $P(x)$ . Confusing $x$ with the number $c$ from the divisor $(x - c)$ breaks the setup.
$c$	The constant that defines which linear factor $(x - c)$ you are dividing by; if the divisor is $(x + 3)$ , then $c = -3$ , not $c = 3$ . Getting the sign wrong flips all the calculations.

When to use it

Use Synthetic Division when you need to divide a polynomial by a linear binomial of the form $(x - c)$ and want a faster shortcut than polynomial long division.

Level

Usually taught in: Algebra II

Worked examples

1. Divide a cubic polynomial by a linear binomial with small integers

Problem

Divide

x^3 + 2x^2 - 5x + 3

(x - 2)

using Synthetic Division.

$\begin{array}{c|cccc} 2 & 1 & 2 & -5 & 3 \\ & & 2 & 8 & 6 \\ \hline & 1 & 4 & 3 & 9 \end{array}$
Set up the synthetic division table with $c = 2$ on the left. Write the coefficients $1, 2, -5, 3$ in the top row.
$1$
Bring down the leading coefficient $1$ to the bottom row—this is the first coefficient of the quotient.
$2 \cdot 1 = 2, \quad 2 + 2 = 4$
Multiply $c = 2$ by the 1 you brought down to get 2, then add it to the next coefficient $2$ to get $4$ .
$2 \cdot 4 = 8, \quad -5 + 8 = 3$
Multiply $2$ by the result $4$ to get $8$ , then add it to the next coefficient $-5$ to get $3$ .
$2 \cdot 3 = 6, \quad 3 + 6 = 9$
Multiply $2$ by $3$ to get $6$ , then add it to the last coefficient $3$ to get $9$ , which is the remainder.

Answer: $x^2 + 4x + 3 + \frac{9}{x - 2}$

Synthetic Division compressed all the steps of polynomial long division into one compact table. The numbers in the bottom row are the coefficients of the quotient ( $1, 4, 3$ give $x^2 + 4x + 3$ ) and the remainder ( $9$ ). Since the quotient always has degree one less than the original polynomial, a cubic divided by a linear divisor gives a quadratic.

2. Divide with a negative divisor and nonzero remainder

Problem

Use Synthetic Division to divide

2x^3 - 3x^2 + x - 7

(x + 1)

$\begin{array}{c|cccc} -1 & 2 & -3 & 1 & -7 \\ & & -2 & 5 & -6 \\ \hline & 2 & -5 & 6 & -13 \end{array}$
Rewrite $(x + 1)$ as $(x - (-1))$ to see that $c = -1$ , not $c = 1$ . Write the coefficients $2, -3, 1, -7$ across the top row.
$2$
Bring down the leading coefficient $2$ —this is the first coefficient of the quotient.
$(-1) \cdot 2 = -2, \quad -3 + (-2) = -5$
Multiply $c = -1$ by $2$ to get $-2$ , then add it to the next coefficient $-3$ to get $-5$ .
$(-1) \cdot (-5) = 5, \quad 1 + 5 = 6$
Multiply $-1$ by $-5$ to get $5$ , then add it to the next coefficient $1$ to get $6$ .
$(-1) \cdot 6 = -6, \quad -7 + (-6) = -13$
Multiply $-1$ by $6$ to get $-6$ , then add it to the last coefficient $-7$ to get $-13$ , the remainder.

Answer: $2x^2 - 5x + 6 - \frac{13}{x + 1}$

The negative divisor changes all the multiplication steps: every product has the opposite sign compared to a positive divisor. The remainder $-13$ is not zero, so $(x + 1)$ is not a factor of the original polynomial. Notice that the quotient still has degree one less than the original cubic.

3. Evaluate polynomial elevation along a hill

Problem

A cyclist climbs a hill whose elevation is modeled by

h(x) = x^3 - 4x^2 + 5x + 2

meters, where

x

is the horizontal distance in kilometers. Use Synthetic Division to divide by

(x - 3)

, which will reveal the elevation at

x = 3

km.

$\begin{array}{c|cccc} 3 & 1 & -4 & 5 & 2 \\ & & 3 & -3 & 6 \\ \hline & 1 & -1 & 2 & 8 \end{array}$
Set up with $c = 3$ . Write the coefficients of $h(x)$ as $1, -4, 5, 2$ in the top row.
$1$
Bring down the leading coefficient $1$ .
$3 \cdot 1 = 3, \quad -4 + 3 = -1$
Multiply $3$ by $1$ to get $3$ , then add to the next coefficient $-4$ to get $-1$ .
$3 \cdot (-1) = -3, \quad 5 + (-3) = 2$
Multiply $3$ by $-1$ to get $-3$ , then add to the next coefficient $5$ to get $2$ .
$3 \cdot 2 = 6, \quad 2 + 6 = 8$
Multiply $3$ by $2$ to get $6$ , then add to the last coefficient $2$ to get $8$ —this remainder is the elevation at $x = 3$ .

Answer: $x^2 - x + 2 + \frac{8}{x - 3}, \text{ or } h(x) = (x - 3)(x^2 - x + 2) + 8$

By the Remainder Theorem, the remainder when dividing $h(x)$ by $(x - 3)$ equals $h(3) = 8$ meters—the elevation at exactly $3$ kilometers. Synthetic Division gives this value instantly without substituting $x = 3$ into the original polynomial. The quotient $x^2 - x + 2$ describes the elevation change relative to that point.

Common mistakes

Where Synthetic Division usually goes wrong

The mistake	Why it is wrong	The fix
Writing the wrong value of $c$ when the divisor is $(x + k)$ : using $c = k$ instead of $c = -k$ .	The divisor $(x + k)$ is the same as $(x - (-k))$ , so $c = -k$ . Using the wrong sign reverses every multiplication step in the algorithm.	Always rewrite $(x + k)$ as $(x - (-k))$ before identifying $c$ , or remember: if you see a plus sign in the divisor, the $c$ value is negative.
Treating the quotient as having the wrong degree, such as writing all numbers from the bottom row as coefficients of the quotient.	The quotient has degree exactly one less than the original polynomial. The last number in the bottom row is the remainder, not a coefficient. If you include it as a coefficient, you add an extra degree to the quotient.	Count how many numbers are in the bottom row. If there are $n$ numbers, the quotient is a polynomial of degree $n - 2$ , and the last number is always the remainder.
Omitting a zero coefficient when the original polynomial is missing a term, such as treating $x^3 + 5x - 2$ as having coefficients $1, 5, -2$ instead of $1, 0, 5, -2$ .	Each position in the synthetic division table represents one term, in order by degree. Skipping a missing term misaligns all the products and additions that follow.	Before starting synthetic division, write the polynomial with all terms in descending degree order, including $0$ coefficients for any missing powers. For $x^3 + 5x - 2$ , write $x^3 + 0x^2 + 5x - 2$ , giving coefficients $1, 0, 5, -2$ .

Tips and when to use something else

Synthetic Division only works for divisors of the form $(x - c)$ where the leading coefficient of $x$ is $1$ . For other linear divisors like $(2x - 1)$ or $(3x + 2)$ , use Polynomial Long Division instead.
The Remainder Theorem says that when you divide $P(x)$ by $(x - c)$ , the remainder equals $P(c)$ —the value of the polynomial at $x = c$ . You can verify your answer by substituting: calculate $P(c)$ and confirm it matches the remainder you found.
If the remainder is zero, then $(x - c)$ is a factor of $P(x)$ , which means $c$ is a root of the polynomial. Use this to find and verify roots quickly.
Always include zero coefficients for missing powers. For $x^4 - 8$ (which is missing $x^3$ , $x^2$ , and $x$ terms), use coefficients $1, 0, 0, 0, -8$ —do not skip directly to the constant term.

Related formulas

Polynomial Long DivisionPolynomial long division breaks a fraction into quotient and remainder, letting you simplify rational expressions or test polynomial divisibility.Remainder TheoremThe Remainder Theorem lets you find the remainder when a polynomial is divided by (x - c) by simply evaluating the polynomial at c.Factor TheoremThe Factor Theorem tells you when a linear binomial divides a polynomial: (x - c) is a factor of P(x) exactly when P(c) = 0.Rational Root TheoremThe Rational Root Theorem tells you which rational numbers could possibly be roots of a polynomial with integer coefficients.Factoring TrinomialsBreak a trinomial into two binomials by finding two numbers that add to b and multiply to c; use when solving equations or simplifying.

Frequently asked questions

Why is Synthetic Division faster than Polynomial Long Division?: Synthetic Division compresses the algorithm into a single compact table and works with only the coefficients, not the variable $x$ . You avoid writing the intermediate quotient and remainder at each step, reducing both the writing and the chance of arithmetic errors.
Can I use Synthetic Division if the divisor has a leading coefficient other than 1?: No, Synthetic Division only works for divisors of the form $(x - c)$ . If your divisor is $(2x - 1)$ or $(3x + 2)$ , factor out the leading coefficient to rewrite it as $2(x - \frac{1}{2})$ or $3(x - \frac{2}{3})$ , then either adjust the final quotient or use Polynomial Long Division instead.
What does the last number in the bottom row of the synthetic division table represent?: By the Remainder Theorem, that last number is the remainder, and it also equals $P(c)$ —the value of the polynomial at $x = c$ . If the remainder is zero, then $(x - c)$ is a factor of $P(x)$ .
What if my polynomial is missing terms, like $x^3 - 2$ ?: Write the coefficients including zeros for missing powers: $x^3 + 0x^2 + 0x - 2$ has coefficients $1, 0, 0, -2$ . If you skip the zeros, the products and additions will not line up correctly with the original polynomial's terms.

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Reviewed 2026-09-18