Remainder Theorem and Factor Theorem

Dividing a polynomial by (x, c) and getting a remainder is the same as plugging x = c into the polynomial itself. That single fact, the remainder theorem, collapses two separate test problems into one quick calculation. Here is how to apply both theorems and use synthetic division to get the answer in under 30 seconds.

Any polynomial P(x) divided by the linear binomial (x, c) fits this identity exactly. P(x) is your original polynomial, (x, c) is the divisor, Q(x) is the quotient (degree one less than P(x)), and R is a constant remainder. This is the polynomial version of 17 ÷ 5 = 3 remainder 2, rewritten as 17 = 5 × 3 + 2. The identity holds for any polynomial degree and any real or complex value of c.

If P(x) has degree n, then Q(x) has degree n, 1. The product (x, c) × Q(x) gives degree n, matching P(x). The remainder R is a constant, degree 0, so the equation balances. Missing terms in P(x) (like a missing x² term) must be entered as zero coefficients in synthetic division, or the identity fails.

The remainder theorem states that when you divide P(x) by (x, c), the remainder R equals P(c). The proof takes two lines. From the division identity, P(x) = (x, c)Q(x) + R. Substitute x = c: P(c) = (c, c)Q(c) + R = 0 × Q(c) + R = R. That is it. The remainder is the polynomial's value at x = c.

This means you never have to perform full division just to evaluate a polynomial at a point. Synthetic division, or synthetic substitution, the same algorithm, gives you P(c) directly as the last number in the bottom row. OpenStax College Algebra 2e section 5.5 (Zeros of Polynomial Functions: Remainder and Factor Theorems) uses this as the foundation for finding zeros.

The factor theorem is a direct corollary: if the remainder R is zero, then (x, c) is a factor of P(x). In other words, c is a root of the polynomial. The theorem says (x, c) is a factor of P(x) if and only if P(c) = 0.

When you run synthetic division and get a remainder of zero, the bottom row (excluding the last number) gives the coefficients of Q(x). You can then factor Q(x) further using the rational root theorem or quadratic methods. A non-zero remainder means (x, c) is not a factor, and the remainder itself is exactly P(c).

Synthetic division is the fastest method for both dividing by (x, c) and evaluating P(c). The algorithm works only for monic linear divisors (x, c). For a divisor like 2x, 3, you must first rewrite it as 2(x, 3/2) and then divide the resulting quotient by 2. Most textbooks omit this step; the site calculator supports (x, c) form only.

Setting Up the Table

Write the coefficients of P(x) in order, including zero for any missing term. For example, for P(x) = 2x³ + 5x², 3x + 7 and divisor (x, 2), the coefficient row is 2, 5,, 3, 7. Bring down the leading coefficient (2) below the line. Multiply by c (2), add to the next coefficient (5): 2 × 2 = 4, 4 + 5 = 9. Repeat: 9 × 2 = 18, 18 + (, 3) = 15; 15 × 2 = 30, 30 + 7 = 37. The bottom row is 2, 9, 15, 37.

Reading the Results

The last number (37) is the remainder R and equals P(2). The numbers before it (2, 9, 15) are coefficients of the quotient Q(x) = 2x² + 9x + 15. The identity P(x) = (x, 2)(2x² + 9x + 15) + 37 holds. If the remainder had been zero, (x, 2) would be a factor.

The Most Common Error: Wrong Sign for c

Dividing by (x + 3) means c =, 3, not +3. The divisor is (x, c), so x + 3 = x, (, 3). Entering +3 instead of, 3 gives a wrong remainder and quotient. Always solve x, c = 0: for (x + 3), c =, 3.

Problem: Evaluate P(x) = x³ + 6x² + 11x + 6 at x = 1 using synthetic division.

Coefficients: 1, 6, 11, 6. c = 1. Bring down 1. Multiply by 1: 1, add to 6: 7. Multiply 7 by 1: 7, add to 11: 18. Multiply 18 by 1: 18, add to 6: 24. Remainder = 24. So P(1) = 24, and (x - 1) is not a factor.

Problem: Is (x, 2) a factor of P(x) = 2x³, 7x² + 3x + 10?

Coefficients: 2,, 7, 3, 10. c = 2. Bring down 2. Multiply by 2: 4, add to, 7:, 3. Multiply, 3 by 2:, 6, add to 3:, 3. Multiply, 3 by 2:, 6, add to 10: 4. Remainder = 4, not zero. (x, 2) is not a factor. P(2) = 4.

Problem: Find k so that (x, 2) is a factor of P(x) = x³, 5x² + kx, 8.

For (x, 2) to be a factor, P(2) must equal 0. Substitute x = 2: 8-20 + 2k, 8 = 0 →, 20 + 2k = 0 → k = 10. Synthetic division with k = 10 gives remainder 0, confirming (x, 2) is a factor.

Problem: Find k so that (x + 1) is a factor of P(x) = 2x³ + kx² + 3x + 5.

Divisor is (x + 1), so c = -1. Set P(-1) = 0: 2(-1)³ + k(-1)² + 3(-1) + 5 = -2 + k - 3 + 5 = k + 0 = 0 → k = 0. With k = 0, synthetic division with c = -1 yields remainder 0.

When you use a synthetic division calculator, the output shows the bottom row of the table. The last number is always the remainder R. The numbers before it are the coefficients of Q(x), in order from highest degree to constant term.

If the remainder is zero, the divisor (x, c) is a factor and c is a root. You can then use the quotient coefficients to write Q(x) and continue factoring. If the remainder is non-zero, it equals P(c). The quotient degree should be exactly one less than the original polynomial's degree. If it is smaller, check for missing terms you forgot to include as zero coefficients.

OpenStax College Algebra 2e section 5.5 covers these interpretations in detail. For the rational root theorem and synthetic division practice problems, those topics receive separate treatment, focus here on applying the remainder and factor theorems to test problems.

The remainder theorem and factor theorem with synthetic division suit Algebra 2 students who need to factor polynomials and find roots quickly for tests, precalculus students evaluating polynomials for graphs, and college algebra students checking rational root candidates efficiently. Teachers also use it to verify student work or generate examples.

Skip this method if you need to divide by a quadratic or higher-degree divisor, use polynomial long division instead. Also skip if you have not yet mastered polynomial long division; synthetic division is a shortcut, not a replacement. The single thing that most often goes wrong is entering the wrong sign for c. Double-check it every time.