What Is Synthetic Division?
Synthetic division is a shorthand method for dividing a polynomial by a linear factor of the form (x − c). It achieves the same result as polynomial long division but is much faster because you only work with coefficients — no variables, no rewriting each term.
Synthetic division is especially useful for: (1) testing whether a given value is a root of a polynomial, (2) factoring a polynomial once you know one root, (3) evaluating polynomials quickly using the remainder theorem, and (4) simplifying polynomial division in pre-calculus and calculus.
When the method applies
The method only works for linear divisors — that is, binomials of the form (x − c). If your divisor is anything else (a quadratic, a linear term with a coefficient other than 1, etc.), you'll need polynomial long division or another technique.
For divisors like (2x − 3), you can factor out the leading coefficient first and then adjust — but for simplicity, stick to linear monic divisors when learning synthetic division.
Synthetic Division vs. Long Division
Polynomial long division works for any divisor (linear, quadratic, or higher). It's the general-purpose method and mirrors the long division you learned for numbers in elementary school. It's slow, error-prone, and takes up a lot of paper, but it always works.
Synthetic division only works for divisors of the form (x − c), but it's dramatically faster. A problem that takes 15 lines of long division can often be done in 2 lines with synthetic division.
When to use which
| Method | Use When | Examples |
|---|---|---|
| Synthetic division | Dividing by a linear term | (x − 2), (x + 5), (x − 1/3) |
| Long division | Dividing by quadratics or higher-degree polynomials | (x² + 1), (x² − 3x + 2) |
On exams, synthetic division is a massive time-saver whenever it applies. Graders also find synthetic division easier to follow, which can help you get partial credit.
We have a complete guide to long division of polynomials for the cases where synthetic division doesn't apply.
The Synthetic Division Method — Step by Step
To divide a polynomial P(x) by (x − c), follow these steps.
- Write the coefficients in order
List the coefficients of
P(x)in descending order of powers. Use0for any missing terms. - Write c on the left
The value
cis the number that makes(x − c) = 0. For divisor(x − 3), usec = 3. For(x + 2), usec = −2. - Bring down the first coefficient
Copy the leading coefficient below the line.
- Multiply and add
Multiply the brought-down value by
c, write the result below the next coefficient, and add. Write the sum below the line. Repeat across all coefficients. - Read the result
The numbers below the line (except the last) are the coefficients of the quotient, in descending order. The quotient has degree one less than
P(x). The last number is the remainder.
1, −4, 5, −2; divisor root is c = 2. The next sections show the full walkthrough.Worked Example: Divide x³ − 4x² + 5x − 2 by (x − 2)
Setup: Coefficients: 1, −4, 5, −2. Divisor: (x − 2), so c = 2. Write this on the left.
Step 1 — Bring down 1: The bottom row starts with 1.
Step 2 — Multiply and add: 1 × 2 = 2. Write 2 under −4. Add: −4 + 2 = −2.
Step 3 — Repeat: −2 × 2 = −4. Write −4 under 5. Add: 5 + (−4) = 1.
Step 4 — Repeat: 1 × 2 = 2. Write 2 under −2. Add: −2 + 2 = 0.
Bottom row: 1, −2, 1, 0.
Read the result: The first three numbers (1, −2, 1) are the coefficients of the quotient, in descending order. Since we started with a degree-3 polynomial, the quotient has degree 2: x² − 2x + 1. The last number (0) is the remainder.
(x³ − 4x² + 5x − 2) ÷ (x − 2) = x² − 2x + 1Since the remainder is zero, (x − 2) is a factor of the original polynomial, and x = 2 is a root. In fact, x² − 2x + 1 = (x − 1)², so P(x) = (x − 2)(x − 1)².
Worked Example: Divide 2x⁴ − 3x³ + 0x² − 5 by (x + 1)
Setup: The polynomial has a missing x² term, so include a 0 coefficient. Coefficients: 2, −3, 0, 0, −5. (We need coefficients for x⁴, x³, x², x¹, x⁰ — five in total.) Divisor: (x + 1) = (x − (−1)), so c = −1.
Walkthrough: Bring down 2. 2 × (−1) = −2. −3 + (−2) = −5.
−5 × (−1) = 5. 0 + 5 = 5.
5 × (−1) = −5. 0 + (−5) = −5.
−5 × (−1) = 5. −5 + 5 = 0.
Bottom row: 2, −5, 5, −5, 0.
Result: Quotient coefficients: 2, −5, 5, −5. Since we started with degree 4, the quotient has degree 3: 2x³ − 5x² + 5x − 5. Remainder: 0.
(2x⁴ − 3x³ − 5) ÷ (x + 1) = 2x³ − 5x² + 5x − 5Since the remainder is zero, x = −1 is a root of the original polynomial.
Note how the missing x² term absolutely requires a 0 coefficient. Skipping it is the most common synthetic division mistake.
The Remainder Theorem and Finding Roots
P(x) by (x − c), the remainder equals P(c). In other words, dividing by (x − c) gives you the value of the polynomial at c as a free byproduct.This makes synthetic division a powerful tool for evaluating polynomials — especially large ones. Instead of plugging in c and computing powers, you just run synthetic division and read the last number.
Setup: coefficients 3, −2, 1, 5, −7, c = 2.
Walkthrough: Bring down 3. 3 × 2 = 6. −2 + 6 = 4.
4 × 2 = 8. 1 + 8 = 9.
9 × 2 = 18. 5 + 18 = 23.
23 × 2 = 46. −7 + 46 = 39.
P(2) = 39(x − c) is a factor of P(x) if and only if P(c) = 0. So to find roots, test candidates by synthetic division; any candidate that produces remainder 0 is a root, and the quotient gives the remaining factor.Combined with the rational root theorem (rational roots of a polynomial are of the form p/q where p divides the constant term and q divides the leading coefficient), synthetic division becomes a systematic way to factor polynomials.
For tricky factoring problems, scan them with Solver AI — it will identify rational root candidates, run synthetic division, and show every step.