Synthetic Division: Worked Examples & Remainder/Factor Theorems

Q: What is synthetic division?

Synthetic division is a shorthand for dividing a polynomial P(x) by a linear divisor x − k. It works only with the coefficients, producing a quotient Q(x) of degree deg(P) − 1 and a constant remainder R, satisfying P(x) = (x − k)·Q(x) + R.

Q: What is the Remainder Theorem?

The Remainder Theorem states that when a polynomial P(x) is divided by x − k, the remainder R equals P(k). So instead of performing the division, you can just evaluate the polynomial at k.

Q: What is the Factor Theorem?

The Factor Theorem states that x − k is a factor of P(x) if and only if P(k) = 0. In other words, dividing by x − k yields a remainder of zero exactly when k is a zero of the polynomial.

Q: When can synthetic division be used?

Synthetic division requires the divisor to be a linear binomial of the form x − k with leading coefficient 1. For divisors like 2x + 3, divide by x + 3/2 first, then divide the quotient by 2. For higher-degree divisors, use polynomial long division.

Overview

Synthetic division is a streamlined method of dividing polynomials by a linear factor \( x - k \). This guide covers the process, the Remainder Theorem, and the Factor Theorem, with worked examples and eight practice problems.

An online synthetic division calculator is also available for checking your work.

Division Algorithm for Polynomials

For any polynomials \( P(x) \) and \( D(x) \ne 0 \), there exist unique polynomials \( Q(x) \) and \( R(x) \) with \( \deg(R) < \deg(D) \) such that:

\[ P(x) = Q(x)\,D(x) + R(x) \]

or equivalently:

\[ \frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)} \]

If \( R(x) = 0 \), then \( D(x) \) is a factor of \( P(x) \).

Steps of Synthetic Division

Synthetic division applies when the divisor has the form \( x - k \). The remainder is then a constant. For dividing \( ax^2 + bx + c \) by \( x - k \):

Step 1 — Set up the table. Write the coefficients \( a, b, c \) in descending order (include 0 for any missing powers) and place \( k \) on the outside left.

Figure 1 — Starting table for synthetic division.
Step 2 — Iterate.
1. Bring the first coefficient \( a \) straight down below the line.
2. Multiply it by \( k \); place the product in the next column above the line.
3. Add that column; place the sum below the line.
4. Repeat until all columns are filled.
Figure 2 — Multiply-by-\(k\) and add, column by column.
Step 3 — Read off the result. The numbers below the line (except the last) are the coefficients of \( Q(x) \), which has degree one less than the dividend. The final number is the remainder \( R \).

Example 1 — Divide \( 2x^2 + 6x - 1 \) by \( x - 3 \)

Step-by-step solution

Set up. The divisor is \( x - 3 \), so \( k = 3 \). Coefficients of the dividend: \( 2, 6, -1 \).

Step 1 for Example 1: k = 3 with coefficients 2, 6, −1

Bring down the leading coefficient. Write \( 2 \) below the line.

Step 2 for Example 1: bringing down the leading coefficient 2

Multiply and add (column 2). \( 3 \cdot 2 = 6 \); \( 6 + 6 = 12 \).

Step 3 for Example 1: multiply 2 by 3, add to 6, giving 12

Multiply and add (column 3). \( 3 \cdot 12 = 36 \); \( -1 + 36 = 35 \).

Step 4 for Example 1: multiply 12 by 3, add to −1, giving 35

Read off the result.

\[ Q(x) = 2x + 12, \qquad R = 35 \] \[ \frac{2x^2 + 6x - 1}{x - 3} = 2x + 12 + \frac{35}{x - 3} \]

Final synthetic division table for Example 1

Check: \( (x-3)(2x+12) + 35 = 2x^2 + 6x - 1 \) ✓

Remainder Theorem

Theorem. If a polynomial \( P(x) \) is divided by \( x - k \), the remainder \( R \) equals \( P(k) \).

Example 2 — Remainder of \( (4x^4 - x^3 + x^2 - 1) \div (x + 1) \)

Solution (two methods)

Method 1 — Remainder Theorem. The divisor is \( x + 1 = x - (-1) \), so \( k = -1 \):

\[ R = P(-1) = 4(1) - (-1) + 1 - 1 = 5 \]

Method 2 — Synthetic division. Coefficients with placeholders: \( 4, -1, 1, 0, -1 \); \( k = -1 \):

Synthetic division of 4x⁴ − x³ + x² − 1 by x + 1, showing remainder 5

The final value is 5 — matching the Remainder Theorem. Answer: \( R = 5 \).

Factor Theorem

Theorem. \( x - k \) is a factor of \( P(x) \) if and only if \( P(k) = 0 \).

Example 3 — Is \( x + 2 \) a factor of \( x^3 + x + 10 \)?

Solution (two methods)

Method 1 — Evaluate \( P(-2) \).

\[ P(-2) = (-2)^3 + (-2) + 10 = -8 - 2 + 10 = 0 \]

Since \( P(-2) = 0 \), \( x + 2 \) is a factor.

Method 2 — Synthetic division with \( k = -2 \) and coefficients \( 1, 0, 1, 10 \) (note the placeholder 0 for \( x^2 \)):

Synthetic division of x³ + x + 10 by x + 2, showing remainder 0

The remainder is 0, so \( x + 2 \) is a factor. The quotient gives the factorization:

\[ x^3 + x + 10 = (x + 2)(x^2 - 2x + 5) \]

Practice Problems with Solutions

Part A — Division of polynomials

(a) Divide \( -4x^4 + 2x^2 - x \) by \( x + 5 \)

Solution

Write all coefficients, including zeros for missing powers: \( -4, 0, 2, -1, 0 \). The divisor is \( x + 5 \), so \( k = -5 \).

Synthetic division of −4x⁴ + 2x² − x by x + 5

Quotient: \( Q(x) = -4x^3 + 20x^2 - 98x + 489 \)

Remainder: \( R = -2445 \)

(b) Divide \( x^5 - 5x^2 - x + 4 \) by \( x - 3 \)

Solution

Coefficients with placeholders: \( 1, 0, 0, -5, -1, 4 \); \( k = 3 \).

Synthetic division of x⁵ − 5x² − x + 4 by x − 3

Quotient: \( Q(x) = x^4 + 3x^3 + 9x^2 + 22x + 65 \)

Remainder: \( R = 199 \)

Part B — Remainder Theorem

(a) Evaluate \( P(2) \) for \( P(x) = -x^3 + 2x^2 + 3x - 8 \)

Solution

By the Remainder Theorem, \( P(2) \) is the remainder of \( P(x) \div (x - 2) \).

Synthetic division of −x³ + 2x² + 3x − 8 by x − 2

\( P(2) = -2 \).

Direct check: \( -8 + 8 + 6 - 8 = -2 \) ✓

(b) Evaluate \( P(0.3) \) for \( P(x) = 0.2x^2 + 0.4x + 0.5 \)

Solution

Apply synthetic division with \( k = 0.3 \):

Synthetic division of 0.2x² + 0.4x + 0.5 by x − 0.3

\( P(0.3) = 0.638 \).

Direct check: \( 0.2(0.09) + 0.4(0.3) + 0.5 = 0.018 + 0.12 + 0.5 = 0.638 \) ✓

(c) Evaluate \( P(1 + \sqrt{2}) \) for \( P(x) = x^3 - 2x^2 + 9 \)

Solution

Apply synthetic division with \( k = 1 + \sqrt{2} \):

Synthetic division of x³ − 2x² + 9 by x − (1 + √2)

\( P(1 + \sqrt{2}) = 10 + \sqrt{2} \).

Direct check: let \( t = 1 + \sqrt{2} \), so \( t^2 = 3 + 2\sqrt{2} \) and \( t^3 = 7 + 5\sqrt{2} \). Then \( P(t) = (7 + 5\sqrt{2}) - 2(3 + 2\sqrt{2}) + 9 = 10 + \sqrt{2} \) ✓

Part C — Factor Theorem

(a) Verify that \( x + 5 \) is a factor of \( P_1(x) = x^3 + 4x^2 - 11x - 30 \)

Solution

The divisor is \( x + 5 = x - (-5) \), so \( k = -5 \). Evaluate \( P_1(-5) \):

\[ P_1(-5) = -125 + 100 + 55 - 30 = 0 \]

Since \( P_1(-5) = 0 \), \( x + 5 \) is a factor.

(b) Verify that \( x + 2 \) and \( x - 3 \) are factors of \( P_2(x) = x^4 - x^3 - 22x^2 + 16x + 96 \)

Solution

Check \( x + 2 \) (\( k = -2 \)):

\[ P_2(-2) = 16 + 8 - 88 - 32 + 96 = 0 \; ✓ \]

Check \( x - 3 \) (\( k = 3 \)):

\[ P_2(3) = 81 - 27 - 198 + 48 + 96 = 0 \; ✓ \]

Both evaluate to zero, so both are factors.

(c) Verify that \( x - 0.4 \) is a factor of \( P_3(x) = x^3 - 0.4x^2 - 0.09x + 0.036 \)

Solution

Here \( k = 0.4 \). Evaluate:

\[ P_3(0.4) = 0.064 - 0.064 - 0.036 + 0.036 = 0 \]

Since \( P_3(0.4) = 0 \), \( x - 0.4 \) is a factor.

(d) Verify that \( x - 1 - \sqrt{3} \) is a factor of \( P_4(x) = x^3 - (5 + 2\sqrt{3})x^2 + (11 + 7\sqrt{3})x - 10 - 6\sqrt{3} \)

Solution

Let \( t = 1 + \sqrt{3} \). Compute the powers:

\[ t^2 = (1 + \sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3} \] \[ t^3 = t \cdot t^2 = (1 + \sqrt{3})(4 + 2\sqrt{3}) = 4 + 2\sqrt{3} + 4\sqrt{3} + 6 = 10 + 6\sqrt{3} \]

Now evaluate each term of \( P_4(t) \):

\( t^3 = 10 + 6\sqrt{3} \)
\( -(5 + 2\sqrt{3}) t^2 = -(5 + 2\sqrt{3})(4 + 2\sqrt{3}) = -(32 + 18\sqrt{3}) \)
\( (11 + 7\sqrt{3}) t = (11 + 7\sqrt{3})(1 + \sqrt{3}) = 32 + 18\sqrt{3} \)
Constant term: \( -10 - 6\sqrt{3} \)

Sum the pieces:

\[ (10 + 6\sqrt{3}) - (32 + 18\sqrt{3}) + (32 + 18\sqrt{3}) - (10 + 6\sqrt{3}) = 0 \; ✓ \]

Since \( P_4(1 + \sqrt{3}) = 0 \), \( x - 1 - \sqrt{3} \) is a factor.

Frequently Asked Questions

What is synthetic division?

Synthetic division is a shorthand for dividing a polynomial \(P(x)\) by \(x - k\). It works only with the coefficients, producing a quotient \(Q(x)\) of degree \(\deg(P) - 1\) and a constant remainder \(R\), with \(P(x) = (x - k)Q(x) + R\).

What is the Remainder Theorem?

When \(P(x)\) is divided by \(x - k\), the remainder \(R = P(k)\). So instead of performing the division, just evaluate the polynomial at \(k\).

What is the Factor Theorem?

\(x - k\) is a factor of \(P(x)\) if and only if \(P(k) = 0\). Dividing by \(x - k\) yields a remainder of zero exactly when \(k\) is a zero of the polynomial.

When can synthetic division be used?

Only when the divisor is \(x - k\) (linear, leading coefficient 1). For a divisor like \(2x + 3\), divide by \(x + 3/2\) first and adjust. For higher-degree divisors, use polynomial long division.

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Last updated: October 17, 2026