How do I handle a divisor like 2x - 3 with synthetic division?

Rewrite the divisor as x - (3/2) and perform synthetic division with c = 3/2. The quotient coefficients you get are for a leading coefficient of 2. Divide every quotient coefficient by 2 to get the correct quotient. The remainder stays the same. The site calculator handles this automatically.

Why does the sign flip when I set up c?

The divisor is x - c. The number c is the root of the equation x - c = 0. If the divisor is x + 5, rewrite it as x - (-5). So c = -5. The sign flips because you are solving x - c = 0, not x + c = 0.

What do I do if the polynomial has a missing term?

Insert a coefficient of 0 for any missing power of x. For example, x³ + 1 becomes x³ + 0x² + 0x + 1. The coefficient row is [1, 0, 0, 1]. Skipping the zeroes misaligns the table and gives wrong results.

How do I know if my synthetic division answer is correct?

Use the Remainder Theorem: the remainder equals P(c). Evaluate the original polynomial at x = c separately. If the two numbers match, your table is correct. The site calculator can verify both the quotient and remainder in seconds.

What is the difference between synthetic division and synthetic substitution?

They are the same algorithm. Synthetic division is the term used when you care about the quotient polynomial. Synthetic substitution is the term used when you only need the remainder (which equals P(c)). The steps are identical.

Synthetic Division Practice Problems With Answers

The Most Common Mistake in Synthetic Division Practice Problems

The most common mistake in synthetic division practice problems is using the wrong sign for c. Dividing by x + 3 requires c = -3, not +3. This error flips every quotient coefficient and the remainder. Correct that single step, and the rest follows cleanly.

Synthetic division practice problems work for any linear binomial divisor of the form x - c. The algorithm uses only the polynomial's coefficients and the value c. It produces the same quotient and remainder as long division but with fewer written steps.

These synthetic division practice problems cover four difficulty levels. Every answer has been verified using the site calculator. Teachers can use these as a synthetic division worksheet generator. Students can treat them as synthetic division exercises with full answer keys.

Level 1: Monic Cubics With Integer c

Monic Cubics With All Terms Present

Start with monic cubics, leading coefficient 1, and a whole-number c. The polynomial has all four terms present.

Problem 1: Divide x³ + 6x² + 11x + 6 by x + 1.
Answer: Quotient x² + 5x + 6, remainder 0. Factored form P(x) = (x + 1)(x² + 5x + 6). The remainder 0 means x + 1 is a factor.

Problem 2: Divide x³ - 4x² + x + 6 by x - 2.
Answer: Quotient x² - 2x - 3, remainder 0. Factored form P(x) = (x - 2)(x² - 2x - 3).

Problem 3: Divide x³ + 2x² - 5x - 6 by x - 3.
Answer: Quotient x² + 5x + 10, remainder 24. Factored form P(x) = (x - 3)(x² + 5x + 10) + 24.

Problem 4: Divide x³ + x² + x + 1 by x - 1.
Answer: Quotient x² + 2x + 3, remainder 4. Factored form P(x) = (x - 1)(x² + 2x + 3) + 4.

Problem 5: Divide x³ + 5x² + 2x - 8 by x + 4.
Answer: Quotient x² + x - 2, remainder 0. Factored form P(x) = (x + 4)(x² + x - 2).

Level 2: Missing Terms and Negative c

Insert Zero Coefficients for Skipped Powers

Polynomials with missing terms require you to insert a zero coefficient for the skipped power. Negative c values test your sign handling.

Problem 6: Divide x⁴ + 0x³ - 5x² + 0x + 4 by x - 2.
Answer: Quotient x³ + 2x² - x - 2, remainder 0. The zero coefficients for x³ and x must be present in the coefficient row.

Problem 7: Divide x³ + 0x² - 3x + 7 by x + 3. Note c = -3.
Answer: Quotient x² - 3x + 6, remainder -11. Factored form P(x) = (x + 3)(x² - 3x + 6) - 11.

Problem 8: Divide x⁴ + 0x³ + 0x² - 16 by x + 2. c = -2.
Answer: Quotient x³ - 2x² + 4x - 8, remainder 0. Factored form P(x) = (x + 2)(x³ - 2x² + 4x - 8).

Problem 9: Divide x³ + 0x² + 0x - 27 by x - 3.
Answer: Quotient x² + 3x + 9, remainder 0. Factored form P(x) = (x - 3)(x² + 3x + 9).

Problem 10: Divide x³ + 0x² + 0x - 64 by x + 4. c = -4.
Answer: Quotient x² - 4x + 16, remainder 0.

Level 3: Fractional c and Non-Monic Divisors

Fractional c Requires Careful Arithmetic

Fractional c values require careful multiply-and-add steps. Non-monic divisors like 2x - 3 add an extra step: after synthetic division with x - (3/2), you must divide the quotient coefficients by 2.

Problem 11: Divide x³ + 3x² - 4x + 2 by x - ½.
Answer: Quotient x² + 3.5x - 2.25, remainder 0.875. Write as x² + (7/2)x - (9/4), remainder 7/8.

Problem 12: Divide x³ - 2x² + x - 1 by x + ⅓. c = -⅓.
Answer: Quotient x² - (7/3)x + (10/9), remainder -37/27.

Problem 13: Divide 2x³ - 5x² + 3x - 7 by 2x - 1. First, rewrite divisor as x - ½. Perform synthetic division with c = ½. Then divide each quotient coefficient by 2.
Answer: Quotient x² - 2x + 0.5, remainder -6.5. Factored form P(x) = (2x - 1)(x² - 2x + 0.5) - 6.5. Verify with the site calculator.

Problem 14: Divide 3x³ + 6x² - 3x + 9 by 3x - 3. Rewrite as x - 1. Synthetic division with c = 1 gives quotient 3x² + 9x + 6, remainder 15. Divide quotient by 3: x² + 3x + 2.
Answer: Quotient x² + 3x + 2, remainder 15.

Problem 15: Divide x³ + 4x² - 7x + 10 by x - 0.4.
Answer: Quotient x² + 4.4x - 5.24, remainder 7.904.

Level 4: Remainder and Factor Theorem Problems

Use the Remainder Theorem to Check Your Work

These problems use the Remainder Theorem, P(c) equals the remainder, and the Factor Theorem, remainder zero means x - c is a factor.

Problem 16: Find P(2) for P(x) = x⁴ - 3x³ + 2x² - x + 5 using synthetic substitution.
Answer: Perform synthetic division with c = 2. Remainder is 3. So P(2) = 3.

Problem 17: Is x + 5 a factor of P(x) = x³ + 7x² + 10x + 50? Use synthetic division with c = -5.
Answer:

Problem 18: Use the Rational Zero Theorem to list all possible rational roots of P(x) = 2x³ - x² + 3x - 6. Then test x = 2 using synthetic division.
Answer: Possible rational roots: ±1, ±2, ±3, ±6, ±½, ±3/2. Synthetic division with c = 2 gives remainder 12. So x = 2 is not a root.

Problem 19: For P(x) = x³ - 2x² - 5x + 6, find P(-3) using synthetic substitution. Then determine if x + 3 is a factor.
Answer: Synthetic division with c = -3 gives remainder -24. P(-3) = -24. Remainder is not 0, so x + 3 is not a factor.

Problem 20: Show that x = 1 is a root of P(x) = x³ - 3x² + 3x - 1. Find the other two roots (they may be complex).
Answer: Synthetic division with c = 1 gives remainder 0. Quotient x² - 2x + 1.

Answers With Full Tables (Collapsible)

Each answer above includes the quotient polynomial, remainder, and factored form P(x) = (x - c)Q(x) + R. The site calculator shows the full synthetic division table for every problem, including the coefficient row, the bring-down step, and each multiply-and-add step. Use it to check your work line by line.

Missing-term problems must show every zero coefficient in the bottom row. For Problem 6, the coefficient row is [1, 0, -5, 0, 4] and c = 2. The bottom row after the algorithm is [1, 2, -1, -2, 0]. The first four numbers are the quotient coefficients: 1, 2, -1, -2. The last number, 0, is the remainder.

When dividing by a non-monic divisor like 2x - 1 (Problem 13), the synthetic division table is done with c = ½. The quotient coefficients from the table are [2, -4, 1]. Dividing each by 2 gives [1, -2, 0.5]. The remainder from the table is -6.5 and is not divided.

Problem Difficulty Summary

Level	Dividend Type	Divisor Type	Special Notes
1	Monic cubic, all terms	x - c, integer c	Basic setup, positive remainder or zero
2	Cubic or quartic with missing terms	x - c, integer or negative c	Zero coefficients required; sign of c
3	Polynomial with fractional c or non-monic divisor	x - c (fraction) or ax - b	Fraction handling; divide quotient by a
4	Any polynomial	x - c (c is root candidate)	Remainder and Factor Theorem application

Who This Subject Suits and Who Should Skip It

Synthetic division suits Algebra 2 students who need to factor polynomials quickly for tests. Precalculus students use it to evaluate polynomials at a point via the Remainder Theorem. College algebra students check rational root candidates efficiently. Teachers use it to verify student work and generate examples. Self-taught learners who find long division tedious can replace it with this faster method for linear divisors.

Skip synthetic division if you need to divide by a quadratic or higher-degree divisor. Use polynomial long division instead. Skip it if you have not yet mastered polynomial long division, synthetic division is a shortcut, not a replacement for understanding the division algorithm. Skip it if you need a general-purpose algebra system; the site calculator is a focused tool, not a computer algebra system.

The single thing that most often goes wrong is entering c with the wrong sign. Dividing by x + 5 means c = -5. That error changes every result. Check it before you start.