Synthetic Division

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 c i + 1 .

In other words: multiply the coefficient just obtained by b , then add the next coefficient of the dividend.

The synthetic division tableau: the bottom row gives the coefficients of <math xmlns= q ( x ) followed by the remainder r ." style="max-width: 100%; height: auto; display: block; margin: 0 auto;">
Figure 1 The synthetic division tableau: the bottom row gives the coefficients of q ( x ) followed by the remainder r .

Summary of synthetic division rules:

  1. c n 1 = a n (first coefficient of quotient equals leading coefficient of dividend)
  2. c i = a i + 1 + b c i + 1 for 0 i n 2 (each subsequent coefficient)
  3. r = a 0 + b c 0 (remainder)

The general synthetic division tableau for a degree- n dividend is:

b a n a n 1 a n 2 a 1 a 0 + c n 1 b c n 2 b c 1 b c 0 b c n 1 c n 2 c n 3 c 0 r

Worked Examples

Example 1.

Divide 3 x 4 5 x 3 4 x 2 + 3 x 2 by x 2 .

Solution Here b = 2 and the coefficients of the dividend are $3, -5, -4, 3, -2$. 2 3 5 4 3 2 + 6 2 4 2 3 1 2 1 4 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 ) = 3 x 3 + x 2 2 x 1 , r = 4. Therefore: 3 x 4 5 x 3 4 x 2 + 3 x 2 = ( x 2 ) ( 3 x 3 + x 2 2 x 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 + 0 x 3 + 0 x 2 + 0 x 1 , so use coefficients $1, 0, 0, 0, -1$.

Example 2.

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 $1, 0, 0, 0, -1$. 1 1 0 0 0 1 + 1 1 1 1 1 1 1 1 0 The remainder is $0$, confirming that x + 1 divides x 4 1 exactly. q ( x ) = x 3 x 2 + x 1 , r = 0.

Dividing by α x β

Synthetic division in its standard form requires a divisor with leading coefficient 1. When the divisor is α x β , rewrite it as α ( x β / α ) , perform synthetic division by x β / α , then adjust the quotient:

  1. Rewrite the divisor as α ( x β α ) .
  2. Divide the polynomial synthetically by x β α , obtaining q ( x ) and r .
  3. The quotient corresponding to divisor α x β is q ( x ) α and the remainder is r .
Why this technique works If f ( x ) = q ( x ) ( x β / α ) + r , multiply and divide by α : f ( x ) = q ( x ) α α ( x β α ) + r = q ( x ) α ( α x β ) + r . So q ( x ) / α is the quotient and r is the remainder when dividing by α x β .
Example 3.

Divide 3 x 3 11 x 2 + 18 x 3 by 3 x 2 .

Solution Write 3 x 2 = 3 ( x 2 / 3 ) and divide synthetically by x 2 / 3 (so b = 2 / 3 ): 2 3 3 11 18 3 + 2 6 8 3 9 12 5 The synthetic quotient is 3 x 2 9 x + 12 and the remainder is $5$. Dividing the quotient by α = 3 : q ( x ) = 3 x 2 9 x + 12 3 = x 2 3 x + 4. Therefore: 3 x 3 11 x 2 + 18 x 3 = ( 3 x 2 ) ( x 2 3 x + 4 ) + 5.

Frequently Asked Questions

When can I use synthetic division? Synthetic division requires the divisor to be a linear polynomial (or reducible to one after factoring out a scalar). It works for divisors of the form x b , x + b (use b ), or α x β (adjust as described above). For quadratic or higher-degree divisors, use long division or undetermined coefficients.
What if I forget a zero coefficient? You will get the wrong quotient and remainder. Always check that the dividend has a coefficient for every power from degree n down to degree 0 before starting synthetic division. Insert explicit zeros for any missing powers.
What does a remainder of 0 tell me? A remainder of 0 means that x b divides f ( x ) exactly, i.e., x b is a factor of f ( x ) . Equivalently, b is a root of f : f ( b ) = 0 . This is the Factor Theorem.