Algebra
How to Do Synthetic Division: Step-by-Step Examples
Divide polynomials by linear factors quickly, handle missing terms and remainders, and use the remainder theorem to check your work
By AI Picture Answer Education Team · · Updated September 25, 2026
What is synthetic division?
Synthetic division is a shortcut for dividing a polynomial by a linear factor of the form (x − r). Instead of writing out every variable like polynomial long division, you work only with the coefficients. The result is the same quotient and remainder in a fraction of the time, which makes synthetic division the go-to method for factoring cubics, testing possible roots, and evaluating polynomials.
As you work through the examples below, you can confirm each answer with the synthetic division calculator, which shows every multiply-and-add step.
Quick definition:
To divide by (x − r), place r in the box, bring down the first coefficient, then repeatedly multiply by r and add to the next coefficient.
When can you use synthetic division?
Standard synthetic division works when the divisor is a linear binomial with x to the first power, such as x − 3 or x + 5. Watch the sign: dividing by x − 3 uses r = 3, while dividing by x + 5 uses r = −5, because x + 5 = x − (−5). If the divisor is quadratic, like x² + 1, use long division instead.
How to do synthetic division in 5 steps
- Write the polynomial in standard form (highest power first) and list its coefficients. Insert a 0 for every missing power.
- Find r by setting the divisor equal to zero. For x − r, the value is r.
- Bring down the first coefficient unchanged.
- Multiply and add. Multiply the number you just wrote by r, place the product under the next coefficient, and add. Repeat until you reach the last column.
- Read the answer. The last number is the remainder. The other numbers are the coefficients of the quotient, which has degree one less than the original polynomial.
Worked examples
Example 1: A cubic with zero remainder
Divide x³ − 6x² + 11x − 6 by (x − 1).
Setup: The coefficients are 1, −6, 11, −6. The divisor is x − 1, so r = 1.
| 1 | 1 | −6 | 11 | −6 |
| 1 | −5 | 6 | ||
| 1 | −5 | 6 | 0 |
Step 1: Bring down the 1.
Step 2: 1 × 1 = 1. Add to −6: −6 + 1 = −5.
Step 3: −5 × 1 = −5. Add to 11: 11 + (−5) = 6.
Step 4: 6 × 1 = 6. Add to −6: −6 + 6 = 0.
Answer: Quotient x² − 5x + 6, remainder 0.
Check: (x − 1)(x² − 5x + 6) = x³ − 5x² + 6x − x² + 5x − 6 = x³ − 6x² + 11x − 6. Because the remainder is 0, x − 1 is a factor, and x² − 5x + 6 = (x − 2)(x − 3). So the full factorization is (x − 1)(x − 2)(x − 3).
Example 2: Missing terms need zero placeholders
Divide x³ − 8 by (x − 2).
Setup: Write x³ − 8 as x³ + 0x² + 0x − 8. The coefficients are 1, 0, 0, −8, and r = 2.
| 2 | 1 | 0 | 0 | −8 |
| 2 | 4 | 8 | ||
| 1 | 2 | 4 | 0 |
Step 1: Bring down the 1.
Step 2: 1 × 2 = 2. Add to 0: 2.
Step 3: 2 × 2 = 4. Add to 0: 4.
Step 4: 4 × 2 = 8. Add to −8: 0.
Answer: Quotient x² + 2x + 4, remainder 0.
Check: (x − 2)(x² + 2x + 4) = x³ + 2x² + 4x − 2x² − 4x − 8 = x³ − 8. This matches the difference of cubes formula.
Example 3: A nonzero remainder and a negative r
Divide 2x⁴ − 3x² + x − 4 by (x + 2).
Setup: The x³ term is missing, so the coefficients are 2, 0, −3, 1, −4. The divisor x + 2 = x − (−2), so r = −2.
| −2 | 2 | 0 | −3 | 1 | −4 |
| −4 | 8 | −10 | 18 | ||
| 2 | −4 | 5 | −9 | 14 |
Step 1: Bring down the 2.
Step 2: 2 × (−2) = −4. Add to 0: −4.
Step 3: −4 × (−2) = 8. Add to −3: 5.
Step 4: 5 × (−2) = −10. Add to 1: −9.
Step 5: −9 × (−2) = 18. Add to −4: 14.
Answer: Quotient 2x³ − 4x² + 5x − 9, remainder 14. Written as a single expression: 2x³ − 4x² + 5x − 9 + 14/(x + 2).
Check with the remainder theorem: f(−2) = 2(16) − 3(4) + (−2) − 4 = 32 − 12 − 2 − 4 = 14. The remainder matches.
Example 4: Dividing by (2x − 1)
Divide 6x³ + 5x² − 2x − 1 by (2x − 1).
Setup: Set 2x − 1 = 0 to get r = 1/2. The coefficients are 6, 5, −2, −1.
| 1/2 | 6 | 5 | −2 | −1 |
| 3 | 4 | 1 | ||
| 6 | 8 | 2 | 0 |
Step 1: Bring down 6. Then 6 × 1/2 = 3, and 5 + 3 = 8.
Step 2: 8 × 1/2 = 4, and −2 + 4 = 2.
Step 3: 2 × 1/2 = 1, and −1 + 1 = 0.
Step 4: This division was by (x − 1/2), which gives 6x² + 8x + 2. Because 2x − 1 = 2(x − 1/2), divide every quotient coefficient by 2.
Answer: Quotient 3x² + 4x + 1, remainder 0.
Check: (2x − 1)(3x² + 4x + 1) = 6x³ + 8x² + 2x − 3x² − 4x − 1 = 6x³ + 5x² − 2x − 1.
The remainder theorem shortcut
The remainder theorem says that when you divide f(x) by (x − r), the remainder equals f(r). That gives you two useful tools. First, a remainder of 0 means r is a root and (x − r) is a factor. Second, synthetic division is often the fastest way to evaluate a polynomial at a number, because it avoids large powers.
Common mistakes to avoid
Watch out for these:
- Using the wrong sign for r. Dividing by x + 4 means r = −4, not 4.
- Skipping zero placeholders. Every missing power needs a 0, or every column after the gap will be wrong.
- Subtracting instead of adding. Long division subtracts; synthetic division always adds.
- Starting the quotient at the wrong degree. The quotient is always one degree lower than the dividend.
- Forgetting to adjust for a leading coefficient. When dividing by ax − b, divide the quotient coefficients by a at the end.
Practice problems
Try these on your own:
- Divide x² + 7x + 10 by (x + 2).
- Divide x³ + 2x² − 5x + 3 by (x − 1).
- Use synthetic division to find f(3) for f(x) = x³ − 4x + 1.
Click to see answers
- x + 5, remainder 0
- x² + 3x − 2, remainder 1
- Coefficients 1, 0, −4, 1 with r = 3 give 1, 3, 5, 16, so f(3) = 16
Frequently asked questions
Can synthetic division be used with any divisor?
No. It works for linear divisors of the form x − r (and ax − b with an adjustment). For quadratic or higher-degree divisors, use polynomial long division.
What does a remainder of zero mean?
A remainder of zero means the divisor is a factor of the polynomial and r is a root, so f(r) = 0.
Why do I need to write zeros for missing terms?
Each column represents one power of x. Leaving out a zero shifts every later coefficient into the wrong column and produces the wrong answer.
Need help with synthetic division?
Once you can divide polynomials, you can factor them and analyze rational functions. Our guide on how to find horizontal asymptotes is a natural next step. For quick answer checks, use the synthetic division calculator or the discriminant calculator for the quadratic you are left with.
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