Synthetic division is a compact numerical shortcut for dividing any polynomial by a linear binomial of the form $x - b$. 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 $b$ 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 $b$; write under the next coefficient; add |
| Result | Bottom row gives coefficients of $q(x)$ (one degree lower), last entry is $r$ |
| For $x + b$ | Use $-b$ in place of $b$ (since $x + b = x - (-b)$) |
Deriving the Algorithm
When we divide a general cubic $a_3 x^3 + a_2 x^2 + a_1 x + a_0$ by $x - b$, the long-division process produces a quotient whose coefficients satisfy a simple pattern. Writing $c_{n-1}$ for the leading coefficient of $q(x)$, each successive coefficient is obtained by:
$c_i = a_{i+1} + b \cdot c_{i+1}.$
In other words: multiply the coefficient just obtained by $b$, then add the next coefficient of the dividend.
Summary of synthetic division rules:
- $c_{n-1} = a_n$ (first coefficient of quotient equals leading coefficient of dividend)
- $c_i = a_{i+1} + b \cdot c_{i+1}$ for $0 \leq i \leq n-2$ (each subsequent coefficient)
- $r = a_0 + b \cdot c_0$ (remainder)
The general synthetic division tableau for a degree-$n$ dividend is:
$ \begin{array}{c|rrrrrr} b & a_n & a_{n-1} & a_{n-2} & \cdots & a_1 & a_0 \\ + & & c_{n-1}b & c_{n-2}b & \cdots & c_1 b & c_0 b \\ \hline & c_{n-1} & c_{n-2} & c_{n-3} & \cdots & c_0 & r \end{array} $
Worked Examples
Divide $3x^4 - 5x^3 - 4x^2 + 3x - 2$ by $x - 2$.
Solution
Here $b = 2$ and the coefficients of the dividend are $3, -5, -4, 3, -2$. $ \begin{array}{c|rrrrr} 2 & 3 & -5 & -4 & 3 & -2 \\ + & & 6 & 2 & -4 & -2 \\ \hline & 3 & 1 & -2 & -1 & -4 \end{array} $ Reading off the bottom row: the quotient coefficients are $3, 1, -2, -1$ (giving a degree-3 quotient) and the remainder is $-4$. $q(x) = 3x^3 + x^2 - 2x - 1, \qquad r = -4.$ Therefore: $3x^4 - 5x^3 - 4x^2 + 3x - 2 = (x-2)(3x^3 + x^2 - 2x - 1) + (-4).$
Important: If the dividend is missing any power of $x$, insert a $0$ coefficient as a placeholder. For example, $x^4 - 1 = x^4 + 0x^3 + 0x^2 + 0x - 1$, so use coefficients $1, 0, 0, 0, -1$.
Divide $x^4 - 1$ by $x + 1$.
Solution
Since $x + 1 = x - (-1)$, we use $b = -1$. The dividend $x^4 - 1$ has missing intermediate terms, so its coefficients are $1, 0, 0, 0, -1$. $ \begin{array}{c|rrrrr} -1 & 1 & 0 & 0 & 0 & -1 \\ + & & -1 & 1 & -1 & 1 \\ \hline & 1 & -1 & 1 & -1 & 0 \end{array} $ The remainder is $0$, confirming that $x + 1$ divides $x^4 - 1$ exactly. $q(x) = x^3 - x^2 + x - 1, \qquad r = 0.$
Dividing by $\alpha x - \beta$
Synthetic division in its standard form requires a divisor with leading coefficient 1. When the divisor is $\alpha x - \beta$, rewrite it as $\alpha(x - \beta/\alpha)$, perform synthetic division by $x - \beta/\alpha$, then adjust the quotient:
- Rewrite the divisor as $\alpha\!\left(x - \dfrac{\beta}{\alpha}\right)$.
- Divide the polynomial synthetically by $x - \dfrac{\beta}{\alpha}$, obtaining $q(x)$ and $r$.
- The quotient corresponding to divisor $\alpha x - \beta$ is $\dfrac{q(x)}{\alpha}$ and the remainder is $r$.
Why this technique works
If $f(x) = q(x)(x - \beta/\alpha) + r$, multiply and divide by $\alpha$: $f(x) = \frac{q(x)}{\alpha} \cdot \alpha\!\left(x - \frac{\beta}{\alpha}\right) + r = \frac{q(x)}{\alpha}(\alpha x - \beta) + r.$ So $q(x)/\alpha$ is the quotient and $r$ is the remainder when dividing by $\alpha x - \beta$.Divide $3x^3 - 11x^2 + 18x - 3$ by $3x - 2$.
