Synthetic division divides a polynomial by a linear expression x − a using only the coefficients, producing the quotient and the remainder in a single row of arithmetic. Dividing x³ + 2x² − 5x + 7 by x − 3 gives the quotient x² + 5x + 10 and a remainder of 37, from three multiply-and-add steps.
The method strips away every x, every plus sign and every subtraction that ordinary division would make you write. What is left is a short table: bring down, multiply by a, add, repeat. The Remainder Theorem Calculator prints the same table for any polynomial.
This guide covers when synthetic division is allowed, how to set up the row (including polynomials with missing terms), the four-step cycle, six worked examples with verification, the zero-remainder case, and eight practice problems with answers.
When You Can Use Synthetic Division
Synthetic division has one hard requirement: the divisor must be linear, that is, of the form x − a.
| Divisor | Can you use synthetic division? | Value of a |
|---|---|---|
| x − 3 | Yes | 3 |
| x + 5 | Yes | −5 |
| x | Yes | 0 |
| 2x − 3 | Yes, with a final rescaling step | 3/2 |
| x² − 4 | No | not applicable |
| x² + x + 1 | No | not applicable |
The dividend can be any degree. Dividing a degree-9 polynomial by x − 2 is nine multiply-and-add steps, and each one is a single multiplication followed by a single addition.
For a quadratic or higher divisor, the shortcut does not exist and you need polynomial long division instead.
Setting Up the Row
Three things go into the setup, and getting them right is most of the work.
1. Write a in the box on the left. Find it by solving divisor = 0. The divisor x − 3 gives a = 3. The divisor x + 4 gives a = −4, because x + 4 is x − (−4). This sign flip is the single biggest source of wrong answers.
2. List the coefficients of f(x) in descending order of power. For 3x³ − 5x² + 2x − 8 the list is 3, −5, 2, −8. Keep the sign attached to each coefficient.
3. Insert a zero for every missing power. The polynomial 2x⁴ − 9x² + 5 has no x³ term and no x term, so its coefficient list is 2, 0, −9, 0, 5. Skipping those zeros shifts every later column and corrupts the whole answer.
A useful check before starting: a degree-n polynomial always produces exactly n + 1 coefficients. Degree 4 means five numbers in the row.
The Four-Step Cycle
- Bring down the leading coefficient unchanged into the bottom row.
- Multiply the number just written in the bottom row by a, and write the product in the middle row of the next column.
- Add the two numbers in that column, writing the sum in the bottom row.
- Repeat steps 2 and 3 until every column is filled.
Notice that synthetic division never subtracts. Long division subtracts at every stage; synthetic division folds that subtraction into the sign of a, which is exactly why the sign of a has to be right.
Reading the Answer
The bottom row holds everything.
- The last number in the bottom row is the remainder.
- Every other number, read left to right, gives the coefficients of the quotient in descending order.
- The quotient’s degree is always one less than the dividend’s degree, because a linear divisor was removed.
So a bottom row of 1, 5, 10, 37 from a cubic dividend means the quotient is x² + 5x + 10 and the remainder is 37.
Example 1: A Cubic Divided by x − 3
Problem. Divide f(x) = x³ + 2x² − 5x + 7 by x − 3.
Solve x − 3 = 0 to get a = 3. Coefficients: 1, 2, −5, 7.
| a = 3 | 1 | 2 | −5 | 7 |
|---|---|---|---|---|
| 3 | 15 | 30 | ||
| 1 | 5 | 10 | 37 |
Step by step:
- Bring down 1.
- 1 × 3 = 3, and 2 + 3 = 5.
- 5 × 3 = 15, and −5 + 15 = 10.
- 10 × 3 = 30, and 7 + 30 = 37.
Answer. Quotient x² + 5x + 10, remainder 37.
Verification. Multiply back using Dividend = Divisor × Quotient + Remainder:
(x − 3)(x² + 5x + 10) + 37
= x³ + 5x² + 10x − 3x² − 15x − 30 + 37
= x³ + 2x² − 5x + 7 ✓Example 2: A Polynomial With Missing Terms
Problem. Divide f(x) = 2x⁴ − 9x² + 5 by x + 2.
The divisor x + 2 gives a = −2. The dividend has no x³ term and no x term, so the coefficient row is 2, 0, −9, 0, 5.
| a = −2 | 2 | 0 | −9 | 0 | 5 |
|---|---|---|---|---|---|
| −4 | 8 | 2 | −4 | ||
| 2 | −4 | −1 | 2 | 1 |
- Bring down 2.
- 2 × (−2) = −4, and 0 + (−4) = −4.
- −4 × (−2) = 8, and −9 + 8 = −1.
- −1 × (−2) = 2, and 0 + 2 = 2.
- 2 × (−2) = −4, and 5 + (−4) = 1.
Answer. Quotient 2x³ − 4x² − x + 2, remainder 1.
Leaving out the two zero placeholders would have produced a row of only three steps and an answer that is wrong in every position.
Example 3: A Remainder of Zero
Problem. Divide f(x) = 3x³ − 5x² + 2x − 8 by x − 2.
a = 2. Coefficients: 3, −5, 2, −8.
| a = 2 | 3 | −5 | 2 | −8 |
|---|---|---|---|---|
| 6 | 2 | 8 | ||
| 3 | 1 | 4 | 0 |
Answer. Quotient 3x² + x + 4, remainder 0.
A remainder of 0 means x − 2 divides f(x) exactly, so x − 2 is a factor and 2 is a root of f(x) = 0. The full statement is:
3x³ − 5x² + 2x − 8 = (x − 2)(3x² + x + 4)This is exactly the situation the Factor Theorem describes, and it is why synthetic division is the standard tool for factoring cubics: once one root is found, the bottom row hands you the quadratic to finish the job.
Example 4: A Negative Value of a
Problem. Divide f(x) = x⁴ + 3x³ − x + 6 by x + 1.
a = −1. The x² term is missing, so the row is 1, 3, 0, −1, 6.
| a = −1 | 1 | 3 | 0 | −1 | 6 |
|---|---|---|---|---|---|
| −1 | −2 | 2 | −1 | ||
| 1 | 2 | −2 | 1 | 5 |
Answer. Quotient x³ + 2x² − 2x + 1, remainder 5.
Verification. Because the remainder equals f(a), check it by substitution: f(−1) = 1 − 3 + 1 + 6 = 5. ✓ That agreement is the Remainder Theorem doing its job, and it is a free check on every synthetic division you run.
Example 5: A Divisor Like 2x − 3
Problem. Divide f(x) = 4x³ − 4x² − 9x + 9 by 2x − 3.
Solve 2x − 3 = 0 to get a = 3/2. Run the cycle as usual:
| a = 3/2 | 4 | −4 | −9 | 9 |
|---|---|---|---|---|
| 6 | 3 | −9 | ||
| 4 | 2 | −6 | 0 |
The bottom row gives 4x² + 2x − 6 with remainder 0, but that quotient belongs to the divisor x − 3/2, not to 2x − 3.
Rescale. Divide the quotient coefficients by the leading coefficient of the real divisor, which is 2:
(4x² + 2x − 6) ÷ 2 = 2x² + x − 3Answer. Quotient 2x² + x − 3, remainder 0.
Verification. (2x − 3)(2x² + x − 3) = 4x³ + 2x² − 6x − 6x² − 3x + 9 = 4x³ − 4x² − 9x + 9. ✓
The remainder never needs rescaling. Only the quotient does.
Example 6: Repeating the Division to Factor Fully
Synthetic division is at its most useful when run twice on the same polynomial.
Problem. Factor f(x) = x³ + 6x² + 11x + 6 completely.
Test a = −1:
| a = −1 | 1 | 6 | 11 | 6 |
|---|---|---|---|---|
| −1 | −5 | −6 | ||
| 1 | 5 | 6 | 0 |
The remainder is 0, so x + 1 is a factor and f(x) = (x + 1)(x² + 5x + 6).
The quadratic factors by inspection: x² + 5x + 6 = (x + 2)(x + 3).
Answer. f(x) = (x + 1)(x + 2)(x + 3), with roots −1, −2 and −3.
Synthetic Division and the Remainder Theorem
The two are the same arithmetic viewed from different ends.
| Remainder Theorem | Synthetic division | |
|---|---|---|
| Input | f(x) and a | f(x) and a |
| Output | The remainder f(a) only | Quotient and remainder |
| Work | One substitution | n multiply-add steps |
| Best for | A quick remainder, a factor check | Factoring, reducing degree |
The final number in a synthetic division row is always f(a). That gives you two independent ways to get the same value, so any disagreement between them tells you an arithmetic slip happened somewhere in the row.
Common Synthetic Division Mistakes
- Using the wrong sign for a. Dividing by x + 6 means a = −6. Solve the divisor for zero every time rather than copying the number you see.
- Omitting zero placeholders. Every missing power needs a 0 in the row. This is the most damaging error, because the answer looks plausible and is entirely wrong.
- Subtracting instead of adding. The cycle adds. Subtraction belongs to long division.
- Writing coefficients in ascending order. The row runs from the highest power down to the constant.
- Reading the whole bottom row as the quotient. The last entry is the remainder and is not a coefficient.
- Forgetting to divide the quotient by the leading coefficient after using a divisor such as 3x − 1.
- Giving the quotient the wrong degree. Dividing a quartic leaves a cubic. If your quotient came out quartic, a column went missing.
Synthetic Division Practice Problems
Divide each polynomial by the given divisor. State the quotient and the remainder.
- x³ − 4x² + x + 6 by x − 2
- 2x³ + 7x² − 5 by x + 1
- x⁴ − 16 by x − 2
- 5x³ − 2x² + x − 3 by x + 2
- x³ − 7x + 6 by x + 3
- 3x⁴ − x³ + 2x − 9 by x − 1
- 4x³ − 4x² − 9x + 9 by 2x − 3
- x⁵ − 2x³ + x − 4 by x + 2
Answers
- Row 1, −2, −3, 0. Quotient x² − 2x − 3, remainder 0. Since the remainder is 0, x − 2 is a factor, and x² − 2x − 3 factors further as (x − 3)(x + 1).
- Coefficients 2, 7, 0, −5. Row 2, 5, −5, 0. Quotient 2x² + 5x − 5, remainder 0.
- Coefficients 1, 0, 0, 0, −16. Row 1, 2, 4, 8, 0. Quotient x³ + 2x² + 4x + 8, remainder 0.
- Row 5, −12, 25, −53. Quotient 5x² − 12x + 25, remainder −53.
- Coefficients 1, 0, −7, 6. Row 1, −3, 2, 0. Quotient x² − 3x + 2, remainder 0.
- Coefficients 3, −1, 0, 2, −9. Row 3, 2, 2, 4, −5. Quotient 3x³ + 2x² + 2x + 4, remainder −5.
- Using a = 3/2 the row is 4, 2, −6, 0. Halving the quotient gives 2x² + x − 3, remainder 0.
- Coefficients 1, 0, −2, 0, 1, −4. Row 1, −2, 2, −4, 9, −22. Quotient x⁴ − 2x³ + 2x² − 4x + 9, remainder −22.
Check any of these by substituting a into the original polynomial. The value you get must equal the remainder.
Synthetic Division FAQ
What is synthetic division used for?
It divides a polynomial by a linear factor x − a quickly, returning both the quotient and the remainder. Its main jobs are evaluating a polynomial, testing whether x − a is a factor, and reducing a cubic or quartic to a quadratic that can then be factored or solved.
Can synthetic division be used with a quadratic divisor?
No. The method depends on the divisor having a single root that can sit in the box, which only happens for a linear divisor. Dividing by x² − 4 or x² + x + 1 requires polynomial long division.
Why do you have to write zeros for missing terms?
Each column in the row stands for one power of x. A missing power still occupies a place value, so it needs a 0 to hold the column. Leaving it out shifts every subsequent coefficient one place and produces an answer of the wrong degree.
What does a remainder of 0 mean in synthetic division?
It means x − a divides the polynomial exactly. So x − a is a factor, a is a root of f(x) = 0, and the bottom row gives the other factor. This is the Factor Theorem stated in table form.
Is synthetic division faster than long division?
For a linear divisor, yes, noticeably. Dividing a quartic by x − 2 takes four multiply-add steps in synthetic division against four full rounds of divide, multiply, subtract and bring down in long division. The gap widens as the degree grows. Long division wins the moment the divisor stops being linear.
How do you handle a divisor like 3x + 6?
Factor out the leading coefficient first: 3x + 6 = 3(x + 2). Run synthetic division with a = −2, then divide the resulting quotient by 3. The remainder from the synthetic row is already correct and needs no adjustment.
Does synthetic division work when a is a fraction or a decimal?
Yes. The arithmetic is identical, only messier. Dividing by 2x − 1 means a = 1/2, and the multiply-and-add steps run on fractions. Keeping the numbers as fractions rather than rounded decimals avoids accumulating error down the row.
How is synthetic division related to finding a remainder in ordinary division?
Both answer the same shape of question, dividend = divisor × quotient + remainder, but on different objects. Whole-number division is covered in how to find the remainder, where the remainder must be smaller than the divisor. In polynomial division the remainder must be of lower degree, which for a linear divisor forces it to be a constant.
