Synthetic division is a compact numerical shortcut for dividing any polynomial by a linear binomial of the form . Instead of writing out all the polynomial notation, we keep only the coefficients and apply a simple multiply-and-add rule. The result is the quotient and remainder in a fraction of the time.
Quick Reference
| Step | Action |
|---|---|
| Setup | Write to the left, then the coefficients of the dividend in order (use 0 for missing powers) |
| Row 1 | Bring down the first coefficient unchanged |
| Repeat | Multiply the last result by ; write under the next coefficient; add |
| Result | Bottom row gives coefficients of (one degree lower), last entry is |
| For | Use in place of (since ) |
Deriving the Algorithm
When we divide a general cubic by , the long-division process produces a quotient whose coefficients satisfy a simple pattern. Writing for the leading coefficient of , each successive coefficient is obtained by:
In other words: multiply the coefficient just obtained by , then add the next coefficient of the dividend.

Summary of synthetic division rules:
- (first coefficient of quotient equals leading coefficient of dividend)
- for (each subsequent coefficient)
- (remainder)
The general synthetic division tableau for a degree- dividend is:
Worked Examples
Divide by .
Solution
Here and the coefficients of the dividend are $3, -5, -4, 3, -2$. Reading off the bottom row: the quotient coefficients are $3, 1, -2, -1$ (giving a degree-3 quotient) and the remainder is . Therefore:
Important: If the dividend is missing any power of , insert a $0$ coefficient as a placeholder. For example, , so use coefficients $1, 0, 0, 0, -1$.
Divide by .
Solution
Since , we use . The dividend has missing intermediate terms, so its coefficients are $1, 0, 0, 0, -1$. The remainder is $0$, confirming that divides exactly.
Dividing by
Synthetic division in its standard form requires a divisor with leading coefficient 1. When the divisor is , rewrite it as , perform synthetic division by , then adjust the quotient:
- Rewrite the divisor as .
- Divide the polynomial synthetically by , obtaining and .
- The quotient corresponding to divisor is and the remainder is .
Why this technique works
If , multiply and divide by : So is the quotient and is the remainder when dividing by .Divide by .