A remainder theorem calculator finds the remainder when a polynomial is divided by a linear factor, without doing full polynomial long division. It is a shortcut that turns a division problem into an evaluation problem, and it is especially useful for checking whether a given value is a root of a polynomial.
The theorem is straightforward. If a polynomial f(x) is divided by (x – a), the remainder is f(a). That means you substitute a into the polynomial and simplify. If the result is zero, then (x – a) is a factor, which is the factor theorem. If the result is not zero, that number is the remainder, and it tells you how far the polynomial is from having that factor.
What this calculator does
This guide uses cubic polynomials, the most common case in school problems, with the form f(x) = px^3 + qx^2 + rx + s. Dividing by (x – a) gives remainder f(a) = p*a^3 + q*a^2 + r*a + s. The calculator evaluates that expression directly, so you can test a divisor value in seconds and then, if needed, confirm by synthetic division on paper.
Pay attention to the sign of a. For divisor (x – 2), a is 2. For divisor (x + 3), the divisor is (x – (-3)), so a is -3. Mixing up that sign is the most common source of a wrong remainder, and it changes the answer even when every other step is correct.
Formula
For f(x) = px^3 + qx^2 + rx + s divided by (x – a), remainder = f(a) = p*a^3 + q*a^2 + r*a + s.
How to use it
- Write the polynomial in descending powers and read off p, q, r, and s, using 0 for any missing term.
- Read a from the divisor: for (x – a), a is the number subtracted; for (x + 3), a is -3.
- Substitute a for x in the polynomial.
- Evaluate powers first, then multiplications, then additions, keeping signs correct.
- If the result is 0, note that (x – a) is a factor. If not, the result is the remainder.
Try the calculator
Worked example
Find the remainder when f(x) = x^3 – 2x^2 + 3x – 5 is divided by (x – 2). Here p = 1, q = -2, r = 3, s = -5, and a = 2. Substituting gives f(2) = 1 x 8 + (-2) x 4 + 3 x 2 – 5 = 8 – 8 + 6 – 5 = 1. The remainder is therefore 1, so (x – 2) is not a factor of this polynomial, but it is close: the polynomial is 1 above a multiple of (x – 2) at x = 2.
A second check with a = 1 gives f(1) = 1 – 2 + 3 – 5 = -3, so dividing by (x – 1) leaves remainder -3. Negative remainders are valid; the remainder is simply the value of the polynomial at a, whatever its sign.
How to check your answer
Confirm with synthetic division for one example. Synthetic division should leave the same final number as f(a).
Test a value that you expect to be a root. If f(a) is 0, you have found a factor and can factor the polynomial further.
Recheck the sign of a before trusting a zero remainder, because a sign error can create a false root.
Common mistakes
- Using a = 3 for divisor (x + 3), when the correct a is -3.
- Skipping a missing term instead of using coefficient 0, which shifts every later coefficient.
- Evaluating powers after multiplication, for example computing 2 x 3 before 2^3, which breaks the order of operations.
Frequently asked questions
What is the remainder theorem in one sentence?
When a polynomial f(x) is divided by (x – a), the remainder is f(a), the value of the polynomial at x = a.
Does a remainder of 0 mean something special?
Yes. It means (x – a) is a factor of the polynomial, which is the factor theorem. That value a is then a root.
Can the remainder be negative?
Yes. The remainder is just f(a), and f(a) can be negative, zero, or positive depending on the polynomial and the value.
