Synthetic Division (and How It Compares)

You have a polynomial to divide, and you know the divisor is simple, something like x - 4. The question is not whether you can do long division, but whether you should. For divisors of the form x - c, synthetic division is the faster, cleaner method. It uses only the coefficients, reduces sign errors, and takes about half the steps. But synthetic division has a hard limit: it only works when the divisor is linear and of the form x - c. If the divisor is x² + 1 or even 2x - 3, synthetic division fails. That is the line. Learn synthetic division for the cases it fits, and use long division for the rest.

Synthetic division is valid only when the divisor is a linear binomial of the form x − c. The constant c can be any real number, positive, negative, integer, fraction, but the leading coefficient of the divisor must be 1. Dividing by x + 5 means c = −5. Dividing by x − ½ means c = ½. That is the only form the algorithm accepts.

The ax − b Workaround

What about a divisor like 2x − 3? The leading coefficient is 2, not 1. Many sources say no, and they are technically correct: you cannot feed 2x − 3 directly into synthetic division. But you can rewrite it. Factor out the 2: 2x − 3 becomes 2(x − ³⁄₂). Divide the dividend by 2 first, then apply synthetic division with c = ³⁄₂. The quotient from that step must be divided again by the factor you pulled out. It adds a step, but it keeps you in synthetic territory. If that feels clunky, use long division. The choice is about speed, not correctness.

Adapting ax − b is a reliable trick, but it is not covered in every Algebra 2 textbook. OpenStax College Algebra 2e (Section 5.4 Dividing Polynomials) presents synthetic division strictly for x − c. If your textbook does the same, and your test problem has 2x − 3, factor first or switch to long division. The failure case is applying synthetic division directly to a non-monic divisor, the algorithm will produce a garbage result.

Synthetic division strips away the variables and works only with the coefficients of the dividend, written in standard form. Every missing term, a term with zero coefficient, must be represented as a zero placeholder. Skipping a placeholder is the most common failure mode and guarantees misalignment.

Step 1: Set Up the Coefficients

Write the coefficients of the dividend in descending order of degree. If the dividend is 3x³ − 5x + 2, note that the x² term is missing. Write it as 3, 0, −5, 2. The missing x² term is 0x². Place these coefficients in a row.

Step 2: Identify c

If the divisor is x − c, c is the number subtracted. For x − 4, c = 4. For x + 7, c = −7. Write c to the left of the coefficients, separated by a vertical line.

Step 3: Bring Down and Multiply

Bring down the first coefficient (the leading coefficient) directly below the line. Multiply it by c and write the result under the next coefficient. Add that column. Multiply the sum by c again, and repeat across the row.

Step 4: Read the Result

The numbers below the line are the coefficients of the quotient, starting one degree lower than the dividend. The last number is the remainder. If the remainder is zero, the Factor Theorem tells you that x − c is a factor of the dividend. If not, the Remainder Theorem says that the remainder equals the value of the original polynomial evaluated at c.

To verify, multiply the quotient by the divisor and add the remainder. The result must equal the original dividend. If it does not, a sign error in subtraction or a missing placeholder is the likely cause.

Work the same problem both ways to see the difference. Divide 2x³ + 5x² − 4x + 1 by x − 2.

Long Division

Set up the long division bracket with the dividend inside and the divisor outside. Divide the leading term: 2x³ divided by x gives 2x². Multiply the divisor by 2x²: 2x³ − 4x². Subtract from the dividend: (2x³ + 5x²) − (2x³ − 4x²) = 9x². Bring down the next term. Repeat: 9x² divided by x gives 9x. Multiply: 9x² − 18x. Subtract: (−4x) − (−18x) = 14x. Bring down the constant 1. Divide: 14x divided by x gives 14. Multiply: 14x − 28. Subtract: 1 − (−28) = 29. The quotient is 2x² + 9x + 14, remainder 29. The result is written as 2x² + 9x + 14 + 29/(x − 2).

Synthetic Division

Write the coefficients of the dividend: 2, 5, −4, 1. Place c = 2 to the left. Bring down the 2. Multiply 2 × 2 = 4, add to 5 gives 9. Multiply 9 × 2 = 18, add to −4 gives 14. Multiply 14 × 2 = 28, add to 1 gives 29. The bottom row reads: 2, 9, 14, 29. The quotient coefficients are 2, 9, 14 (for 2x² + 9x + 14) and the remainder is 29. Same result, about one-third the writing.

The Remainder Theorem states that when a polynomial P(x) is divided by x − c, the remainder equals P(c). In the example above, P(2) = 2(8) + 5(4) − 4(2) + 1 = 16 + 20 − 8 + 1 = 29, matching the remainder. This connection works in both directions: synthetic division gives you the remainder, and evaluating P(c) gives you the same number without doing the division. For quick checks or when you only need the remainder, such as determining if x − c is a factor, evaluate P(c) directly. If P(c) = 0, the remainder is zero, and by the Factor Theorem, x − c is a factor. Synthetic division becomes most useful when you need both the quotient and the remainder.

Long division is the universal algorithm. Use it when the divisor is not of the form x − c. That includes quadratic divisors like x² + x − 1, cubic divisors like x³ − 2, and any divisor with a leading coefficient other than 1 that you cannot or will not adapt. Long division also handles divisors of higher degree than the dividend, in which case the quotient is zero and the remainder is the dividend itself, a case that trips up many students who try to force synthetic division.

Polynomial long division examples with missing terms, fractions, and non-monic divisors are the best practice for mastering alignment. OpenStax College Algebra 2e (Section 5.4) and OpenStax Precalculus 2e (Section 3.7 Rational Functions) both use long division to find oblique asymptotes, a key application in precalculus. When you need the quotient for a rational function and the divisor is not linear, there is no shortcut. Long division is the only option.

The table below summarises when each method applies.

Choose synthetic division when you need both the quotient and the remainder for a linear divisor with leading coefficient 1, and you are comfortable with the setup. It also suits early calculus students integrating rational functions via partial fractions, where a quick quotient is needed before decomposition. Tutors who verify step-by-step work benefit from the speed and reduced risk of sign errors in subtraction.

Skip synthetic division if you only need to multiply or add polynomials, use direct expansion or combining like terms instead. Skip it if you need to divide numbers, use standard arithmetic long division. Skip it if you work with a computer algebra system and do not care about intermediate steps, a CAS will handle any division automatically. The algorithm is a tool for understanding, not a replacement for judgment. It is not faster when it produces a wrong answer.