Synthetic Division Calculator
Perform synthetic division of any polynomial by a linear factor (x−c) — displaying the full synthetic division tableau with every multiplication and addition step shown, the quotient polynomial, remainder, and Remainder Theorem verification. Includes complete polynomial factoring and polynomial long division.
Synthetic Division Tableau
Division Result
Complete Factorization
Long Division Result
Table A: Synthetic Division Steps
| Step | Action | Example (÷(x−3), P=[2,−3,1,−5]) |
|---|---|---|
| 1 | Write c and all coefficients in a row | 3 │ 2 −3 1 −5 |
| 2 | Bring down the leading coefficient as-is | Bottom row: 2 |
| 3 | Multiply bottom value by c, write above next coeff | 2 × 3 = 6 (write above −3) |
| 4 | Add the column: coeff + multiply value | −3 + 6 = 3 |
| 5 | Repeat steps 3–4 for all remaining columns | 3×3=9, 1+9=10; 10×3=30, −5+30=25 |
| 6 | Last number in bottom row = REMAINDER | R = 25 |
| 7 | Other bottom row numbers = quotient coefficients (degree−1) | Q = 2x² + 3x + 10 |
Table B: Remainder Theorem & Factor Theorem
| Theorem | Statement | Practical Use |
|---|---|---|
| Remainder Theorem | P(c) = remainder when P(x) ÷ (x−c) | Evaluate P at a point without expanding |
| Factor Theorem | (x−c) is a factor ↔ P(c) = 0 ↔ R = 0 | Test if c is a root of P(x) |
| Combined use | If synthetic division gives R=0, then c is a root AND (x−c) is a factor AND the quotient is P(x)÷(x−c) | Full factoring step |
Table C: Rational Root Theorem
| Part | Formula | Example: 6x³−x²−4x+1 |
|---|---|---|
| p = factors of constant term | factors of |a₀| | factors of 1: ±1 |
| q = factors of leading coefficient | factors of |aₙ| | factors of 6: ±1, ±2, ±3, ±6 |
| Candidates = ±p/q | all combinations ±p/q | ±1, ±1/2, ±1/3, ±1/6 |
| Test each | Use synthetic division — if R=0, factor found | Test each candidate until R=0 |
Table D: When to Use Synthetic vs Long Division
| Situation | Method | Why |
|---|---|---|
| Divisor is (x−c) — linear, monic | Synthetic division | Fastest, fewest steps |
| Divisor degree ≥ 2 (e.g. x²+1) | Polynomial long division | Synthetic division cannot handle this |
| Need P(c) quickly | Remainder Theorem + synthetic division | Remainder IS P(c) |
| Testing rational roots | Rational Root Theorem + synthetic division | Systematic factoring approach |
| Divisor has leading coeff ≠ 1 | Long division (or adjust synthetic) | Synthetic division requires monic divisor |
Table E: Common Perfect Divisors to Test
| If P(c) = 0 for... | Then... | Factor |
|---|---|---|
| c = 1 | Sum of all coefficients = 0 | (x−1) is a factor |
| c = −1 | Alternating sum of coefficients = 0 | (x+1) is a factor |
| c = 2 | P(2) = 0 | (x−2) is a factor |
| c = −2 | P(−2) = 0 | (x+2) is a factor |
What Is Synthetic Division? — Overview and When to Use It
Synthetic division is a streamlined method for dividing a polynomial P(x) by a linear factor (x−c), using only the numerical coefficients — no variables needed. It is faster than polynomial long division for linear divisors because it reduces the process to a simple pattern of multiplication and addition. This synthetic division calculator performs polynomial division instantly and shows the complete tableau.
The result of synthetic division is: P(x) = (x−c) · Q(x) + R, where Q(x) is the quotient polynomial (degree one less than P) and R is the constant remainder. The Remainder Theorem states that R = P(c) — so synthetic division simultaneously divides AND evaluates the polynomial.
When to use synthetic division: ONLY when the divisor is a monic linear factor (x−c). If the divisor has degree ≥ 2 (like x²+1 or x²−4), use polynomial long division instead. Synthetic division is essentially Horner's method for polynomial evaluation — each step of the algorithm computes exactly what Horner's method computes.
How to Do Synthetic Division — Step-by-Step
The six-step synthetic division procedure creates the tableau format recognized in every algebra textbook:
- Write c from the divisor (x−c) on the left — for (x−3) write 3; for (x+2) write −2
- Write ALL coefficients of P(x) in a row — include ZERO for any missing degree terms
- Bring down the leading coefficient unchanged to the bottom row
- Multiply the bottom row value by c and write above the next coefficient
- Add the column: coefficient + multiply value = next bottom row value
- Repeat steps 4–5; the final bottom row value is the remainder, the rest form the quotient
Full Example: 2x³ − 3x² + x − 5 ÷ (x − 3)
- c = 3 (from x − 3)
- Bring down 2 → bottom row: 2
- 2 × 3 = 6; −3 + 6 = 3
- 3 × 3 = 9; 1 + 9 = 10
- 10 × 3 = 30; −5 + 30 = 25 (REMAINDER)
- Quotient: 2x² + 3x + 10 · Remainder: 25
Critical Warning — Missing Terms (Zero Coefficients): If P(x) = x³ + 4, you MUST write coefficients as [1, 0, 0, 4] — inserting zeros for the missing x² and x terms. Forgetting these zero placeholders produces completely wrong answers. This is the single most common synthetic division error.
The Remainder Theorem — P(c) Equals the Remainder
The Remainder Theorem states: when polynomial P(x) is divided by (x−c), the remainder R equals P(c). This is a powerful result — it means you can evaluate a polynomial at any value x=c simply by performing synthetic division and reading the remainder.
Practical use of the Remainder Theorem: to find P(5) for a high-degree polynomial, just perform synthetic division by (x−5) — the remainder IS P(5), without substituting into the full expression. Our Remainder Theorem calculator shows this verification automatically alongside every synthetic division result.
Remainder Theorem Example: Find P(3) for P(x) = 2x³ − 3x² + x − 5
- Perform synthetic division with c = 3: remainder = 25
- Therefore P(3) = 25 by the Remainder Theorem
- Verify directly: P(3) = 2(27) − 3(9) + 3 − 5 = 54 − 27 + 3 − 5 = 25 ✓
The Factor Theorem — Testing Whether (x−c) Is a Factor
The Factor Theorem is a direct consequence of the Remainder Theorem: (x−c) is a factor of P(x) if and only if P(c) = 0, which means the remainder = 0. This gives a practical test: perform synthetic division; if the remainder is zero, (x−c) is a factor.
Factor Theorem Example: Is (x−1) a factor of x³ − 6x² + 11x − 6?
Remainder = 0 → (x−1) IS a factor. Quotient: x² − 5x + 6 = (x−2)(x−3)
Therefore: x³ − 6x² + 11x − 6 = (x−1)(x−2)(x−3)
Factoring Polynomials Using Synthetic Division — The Rational Root Theorem
Combine the Rational Root Theorem with synthetic division to factor polynomials completely. The Rational Root Theorem states: if P(x) has integer coefficients and p/q (in lowest terms) is a rational root, then p divides the constant term and q divides the leading coefficient.
Method: (1) List candidates ±p/q using the Rational Root Theorem. (2) Test each candidate using synthetic division. (3) When remainder = 0, factor found — use the quotient as the new polynomial. (4) Repeat on the reduced quotient until fully factored.
Full Factoring: x³ − 6x² + 11x − 6
- Rational Root Theorem: candidates = ±1, ±2, ±3, ±6
- Test c=1: synthetic division gives R=0 → (x−1) is a factor
- Quotient: x² − 5x + 6
- Factor quadratic: x² − 5x + 6 = (x−2)(x−3)
- Result: x³ − 6x² + 11x − 6 = (x−1)(x−2)(x−3) · Zeros: x = 1, 2, 3
Missing Terms — The Zero Coefficient Trap
The most common error in synthetic division is forgetting placeholder zeros for missing degree terms. Every degree from the highest down to zero must have a coefficient entry — even if that coefficient is 0.
⚠️ The Zero Coefficient Trap: P(x) = x³ + x + 1 has NO x² term. Coefficients MUST be [1, 0, 1, 1] — NOT [1, 1, 1]. Using [1, 1, 1] is equivalent to dividing x² + x + 1 (a completely different polynomial), giving a totally wrong answer. The zero coefficients represent the missing degree terms and are essential for the tableau to work correctly.
Missing Term Example: x³ + 4 ÷ (x − 2)
WRONG (missing zeros): coefficients [1, 4] → gives completely wrong result
CORRECT: x³ + 0x² + 0x + 4 → coefficients [1, 0, 0, 4]
Quotient: x² + 2x + 4 · Remainder: 12 · So x³+4 = (x−2)(x²+2x+4) + 12
Synthetic Division vs Polynomial Long Division
Both methods produce identical results for linear divisors — synthetic division is simply a compressed version of polynomial long division that omits the variable symbols and only tracks coefficients.
| Feature | Synthetic Division | Polynomial Long Division |
|---|---|---|
| Works for linear divisors (x−c) | ✓ Yes (preferred) | ✓ Yes (slower) |
| Works for quadratic divisors | ✗ No — cannot use | ✓ Yes |
| Speed | Faster — fewer operations | Slower — more steps |
| Requires writing variables | No — coefficients only | Yes — full polynomial notation |
| Evaluates P(c) simultaneously | Yes — remainder = P(c) | No |
Worked Examples — 8 Complete Problems
1. 2x³ − 3x² + x − 5 ÷ (x − 3)
Quotient: 2x²+3x+10 · R: 25 · NOT a factor
2. x³ − 6x² + 11x − 6 ÷ (x − 1) [Factor Theorem check]
R = 0 → (x−1) IS a factor ✓ · Quotient: x²−5x+6 = (x−2)(x−3)
3. x⁴ − 1 ÷ (x − 1) [Note: x⁴+0x³+0x²+0x−1]
Quotient: x³+x²+x+1 · R = 0 → (x−1) IS a factor ✓
4. 3x³+2x²−x+4 ÷ (x+2) → c = −2
Quotient: 3x²−4x+7 · R: −10 · Note: (x+2) means c=−2
5. x³ − 7x + 6 ÷ (x − 2) [Missing x² term → [1,0,−7,6]]
R=0 → (x−2) IS a factor ✓ · Quotient: x²+2x−3 = (x+3)(x−1)
6. x³ + 4 ÷ (x − 2) [Missing terms: 1,0,0,4]
Quotient: x²+2x+4 · R: 12 · x³+4 = (x−2)(x²+2x+4)+12
7. 2x³+3x²−11x−6 ÷ (x+3) → c = −3
R=0 → (x+3) IS a factor ✓ · Quotient: 2x²−3x−2 = (2x+1)(x−2)
8. Find P(2) for P(x) = 3x⁴ − 2x³ + x − 7 using Remainder Theorem
P(2) = remainder = 27 · Verify: 3(16)−2(8)+2−7 = 48−16+2−7 = 27 ✓
Common Mistakes in Synthetic Division
Mistake 1 — Missing Zero Coefficients (Most Common)
- ❌ x³+x+1 entered as [1,1,1] — wrong polynomial entirely
- ✅ x³+0x²+x+1 → [1, 0, 1, 1] — insert zero for every missing degree
Mistake 2 — Wrong Sign for c
- ❌ For divisor (x+2), using c=+2
- ✅ For divisor (x+2) = (x−(−2)), c = −2
- Rule: c is what makes the divisor zero. x+2=0 → x=−2, so c=−2
Mistake 3 — Reading Result Incorrectly
- ❌ Treating the last number as a quotient coefficient
- ✅ The LAST number in the bottom row is the remainder; all others are quotient coefficients (degree one less than dividend)
Mistake 4 — Multiplying Before Adding
- ❌ Computing result[i] × c before adding coefficients[i]
- ✅ Correct order: (result[i−1] × c) + coefficients[i] = result[i]
Mistake 5 — Not Verifying with Remainder Theorem
- ❌ Accepting the answer without checking
- ✅ Always verify: compute P(c) directly and confirm it equals your remainder. P(c) ≠ remainder means an arithmetic error occurred.