Remainder Theorem Calculator

Use this Remainder Theorem Calculator to enter values, adjust options, and review results in a compact responsive workspace.

Results are calculated automatically as you enter data.

Polynomial f(x) Divisor x - a, where a =

Calculations use exact rational arithmetic. Each coefficient and a may use up to 18 significant digits and a decimal exponent from −18 to 18.

Formula and interpretation

Remainder f(a)

Disclaimer: All tools on this site are intended for educational and informational purposes only. Estimate only; not professional advice. Verify units, assumptions, precision, and method before relying on results. See our Terms of Use.

Assumptions and limitations

This Remainder Theorem Calculator is for general educational use. Results are estimates or mathematical outputs based on the inputs and stated assumptions, not professional engineering, safety, financial, legal, or academic advice. Check units, rounding, formulas, and limitations before using results for important work. For professional, safety-sensitive, construction, engineering, official, or high-cost uses, verify results with an appropriate method, current standards, required tolerances, and a qualified professional. Use of this calculator is subject to our Terms of Use.

▼ See explanations and tips below ▼

Related Calculators

What Is the Remainder Theorem?

The remainder theorem is a shortcut for finding the remainder when a polynomial is divided by a linear expression of the form $x - a$. Instead of carrying out a full polynomial division first, you can substitute $a$ into the polynomial.

If the polynomial is $f(x)$, then the remainder after dividing by $x - a$ is:

$$ R = f(a) $$

This means that a division question can become an evaluation question. For example, to find the remainder when $f(x) = x^2 - 4x + 3$ is divided by $x - 1$, evaluate $f(1)$:

$$ f(1) = 1^2 - 4(1) + 3 = 0 $$

So the remainder is $0$. A zero remainder is especially important because it means the divisor goes into the polynomial evenly. In this example, $x - 1$ is a factor of $x^2 - 4x + 3$.

The theorem is useful because it connects three big algebra ideas: polynomial division, function evaluation, and factors. That connection makes it faster to check possible zeros, test factors, and understand how a polynomial can be rewritten as a divisor times a quotient plus a remainder.

Why the Remainder Theorem Matters

Polynomial division can be time-consuming, especially for higher-degree polynomials. The remainder theorem gives a faster path when the divisor is linear and has the form $x - a$.

Students often use it to:

find a remainder without completing long division;
test whether a number is a zero of a polynomial;
decide whether $x - a$ is a factor;
connect synthetic division to function values;
check polynomial division work.

The theorem also helps explain why roots and factors are linked. If $f(a) = 0$, then $a$ is a zero of the polynomial, and $x - a$ is a factor. If $f(a) \ne 0$, then the same division still works, but the leftover amount is the nonzero remainder.

Key Terms to Know

Polynomial: An expression made from constants and nonnegative whole-number powers of a variable, such as $2x^3 - 3x^2 + 4x - 5$.
Dividend: The polynomial being divided.
Divisor: The expression you divide by. For the remainder theorem, the divisor is $x - a$.
Quotient: The polynomial result of the division before the remainder is included.
Remainder: The leftover constant after dividing by $x - a$.
Zero or root: A value $a$ that makes $f(a) = 0$.
Factor: An expression that divides a polynomial with remainder $0$.
Synthetic division: A compact coefficient-based method for dividing by a linear divisor such as $x - a$.
Horner's method: A nested evaluation method that uses repeated multiplication and addition. The running values in synthetic division follow the same basic idea.

How the Remainder Theorem Works

The idea starts with the polynomial division identity. When a polynomial $f(x)$ is divided by $x - a$, the result can be written as:

$$ f(x) = (x - a)Q(x) + R $$

Where:

$Q(x)$ is the quotient polynomial;
$R$ is the remainder;
$x - a$ is the divisor.

Because $x - a$ is a first-degree divisor, the remainder is a constant. Now substitute $x = a$ into the identity:

$$ f(a) = (a - a)Q(a) + R $$

Since $a - a = 0$, the quotient part disappears:

$$ f(a) = 0 \cdot Q(a) + R $$

So:

$$ f(a) = R $$

That is the remainder theorem: the remainder is the polynomial value at $a$.

This also gives the factor theorem. If $f(a) = 0$, then the remainder is $0$, so $x - a$ is a factor of $f(x)$. If $x - a$ is a factor, then dividing by it leaves no remainder, so $f(a) = 0$.

How Synthetic Division Fits In

Synthetic division is a quick way to divide a polynomial by $x - a$ using only coefficients. It produces both the quotient and the remainder.

For example, take:

$$ f(x) = 2x^3 - 3x^2 + 4x - 5 $$

and divide by:

$$ x - 2 $$

Here, $a = 2$. Start with the coefficients:

$$ 2,\ -3,\ 4,\ -5 $$

Synthetic division repeatedly multiplies the current running value by $a$ and adds the next coefficient:

$$ \begin{aligned} \text{Start: }&2 \\ 2 \times 2 + (-3) &= 1 \\ 1 \times 2 + 4 &= 6 \\ 6 \times 2 + (-5) &= 7 \end{aligned} $$

The last running value is the remainder:

$$ R = 7 $$

The earlier running values are the quotient coefficients, so the quotient is:

$$ Q(x) = 2x^2 + x + 6 $$

The division identity is:

$$ 2x^3 - 3x^2 + 4x - 5 = (x - 2)(2x^2 + x + 6) + 7 $$

The same remainder appears if you evaluate the polynomial directly:

$$ f(2) = 2(2)^3 - 3(2)^2 + 4(2) - 5 = 7 $$

Synthetic division therefore gives more than the remainder. It also gives the quotient and a step-by-step check of the division.

Examples of the Remainder Theorem in Practice

Example 1: A Simple Remainder

Find the remainder when:

$$ f(x) = x^2 - 4x + 3 $$

is divided by:

$$ x - 2 $$

The divisor is $x - a$, so $a = 2$. Evaluate $f(2)$:

$$ f(2) = 2^2 - 4(2) + 3 $$

$$ f(2) = 4 - 8 + 3 = -1 $$

The remainder is:

$$ R = -1 $$

That means $x - 2$ is not a factor of $x^2 - 4x + 3$.

Example 2: A Factor Check

Check whether $x - 1$ is a factor of:

$$ f(x) = x^3 - 6x^2 + 11x - 6 $$

Here, $a = 1$. Evaluate:

$$ f(1) = 1^3 - 6(1)^2 + 11(1) - 6 $$

$$ f(1) = 1 - 6 + 11 - 6 = 0 $$

The remainder is $0$, so $x - 1$ is a factor. The value $x = 1$ is also a zero of the polynomial.

Example 3: A Negative Value of $a$

Find the remainder when:

$$ f(x) = x^3 + 8 $$

is divided by:

$$ x + 2 $$

Rewrite the divisor in the form $x - a$:

$$ x + 2 = x - (-2) $$

So $a = -2$. Evaluate:

$$ f(-2) = (-2)^3 + 8 $$

$$ f(-2) = -8 + 8 = 0 $$

The remainder is $0$, so $x + 2$ is a factor. This is a common place for sign mistakes: a divisor written as $x + 2$ corresponds to $a = -2$, not $a = 2$.

How to Interpret the Result

The main result is the remainder $R$. It tells you what is left after dividing the polynomial by $x - a$.

If $R = 0$, then the division is exact:

$$ f(x) = (x - a)Q(x) $$

In that case, $x - a$ is a factor, and $a$ is a zero of the polynomial.

If $R \ne 0$, then the divisor is not a factor:

$$ f(x) = (x - a)Q(x) + R $$

The quotient still matters because it tells you the polynomial part of the division. The remainder tells you the leftover constant.

When a divisor is displayed as $x + c$, it means the entered value of $a$ was negative:

$$ x - (-c) = x + c $$

The synthetic-division table should be read from left to right. Each running value is formed by multiplying the previous running value by $a$ and adding the next coefficient. The final running value is the remainder. The earlier running values become the quotient coefficients.

For nonzero constant polynomials, the quotient is $0$ and the remainder is the constant itself. For example, if $f(x) = 5$, then dividing by $x - 3$ leaves a remainder of $5$.

Common Mistakes and Misconceptions

Entering the whole divisor instead of $a$. If the divisor is $x - 4$, the value of $a$ is $4$. If the divisor is $x + 4$, the value of $a$ is $-4$.

Forgetting that $x + c$ means $a = -c$. The remainder theorem always uses $x - a$. A plus sign in the divisor usually means the value being substituted is negative.

Using a factored expression when an expanded polynomial is needed. A polynomial such as $(x - 1)(x + 2)$ should be expanded first if the input expects expanded terms. In expanded form, it is:

$$ (x - 1)(x + 2) = x^2 + x - 2 $$

Typing fractional coefficients as slash expressions. Slash-style fractions such as $1/2$ are not accepted in the polynomial field. Enter a decimal such as $0.5$; supported decimals are converted to exact reduced fractions and may produce fraction output.

Expecting decimal approximations. Supported decimal inputs are converted to exact reduced fractions, so intermediate values are not approximated and a nonzero remainder is never treated as zero.

Thinking the quotient and remainder are the same thing. The quotient is the polynomial part of the division. The remainder is the leftover constant.

Assuming a nonzero remainder is an error. A nonzero remainder simply means the divisor is not a factor.

When to Use the Remainder Theorem

Use the remainder theorem when you need to divide a polynomial by a linear divisor of the form $x - a$ and you mainly want the remainder.

It is especially useful for:

checking whether a proposed factor is valid;
testing possible zeros of a polynomial;
evaluating a polynomial through synthetic division;
verifying a polynomial division identity;
simplifying work before factoring a higher-degree polynomial;
comparing direct substitution with synthetic-division results.

If the divisor is not linear, such as $x^2 - 1$, the basic remainder theorem in the form $R = f(a)$ does not apply directly. Polynomial long division or a more general division method is needed.

Limitations and Things to Keep in Mind

The remainder theorem applies to division by $x - a$. It does not directly find the remainder for arbitrary divisors such as $x^2 + 1$ or $2x - 3$ unless the problem is first rewritten in a compatible way and handled with the appropriate method.

For this calculator, the polynomial should be a single-variable expression in $x$ written in expanded form. It is intended for expressions such as:

$$ x^2 - 4x + 3 $$

or:

$$ 2x^3 - 3x^2 + 4x - 5 $$

It is not intended for grouped or factored expressions such as:

$$ (x - 1)(x + 2) $$

The polynomial field is numerical, not a full symbolic algebra system. It supports plain integer and decimal coefficients, but it does not support symbolic coefficients, multiple variables, negative exponents, fractional exponents, or slash-style fractions such as $1/2$.

The supported powers are nonnegative integers up to degree $20$. Very large calculations may be too large to evaluate reliably.

The calculator uses exact rational arithmetic for supported decimal inputs. Coefficients and $a$ are converted to reduced fractions before synthetic division, so results are shown as exact integers or fractions such as $-2489/2500$. A divisor is a factor only when the reduced remainder numerator is exactly $0$; a tiny nonzero remainder is not rounded away or treated as zero.

How to Use This Calculator

Enter an expanded polynomial in $x$, such as x^2 - 4x + 3.
Enter the finite value of $a$ for the divisor $x - a$.
If the divisor is written as $x + c$, enter $a = -c$.
Review the remainder result. By the remainder theorem, this value is $f(a)$.
Check the divisor, quotient, factor check, substitution step, synthetic-division table, and division identity to understand the calculation.
If the factor check says the divisor is a factor, the reduced remainder is exactly zero.
Use any built-in examples to see sample inputs, or clear the fields to start again.

Frequently Asked Questions

What does the value of $a$ mean?

The value $a$ is the number in the divisor $x - a$. It is also the number substituted into the polynomial. For example, if the divisor is $x - 5$, then $a = 5$ and the remainder is $f(5)$.

What should I enter for a divisor like $x + 3$?

Rewrite $x + 3$ as $x - (-3)$. That means $a = -3$. Enter -3, not 3.

Is the remainder theorem the same as the factor theorem?

They are closely related, but they answer slightly different questions. The remainder theorem says the remainder after dividing by $x - a$ is $f(a)$. The factor theorem says $x - a$ is a factor exactly when $f(a) = 0$.

Why does a zero remainder prove that $x - a$ is a factor?

If the remainder is zero, the division identity becomes $f(x) = (x - a)Q(x)$. That means $x - a$ multiplies another polynomial to produce $f(x)$, so it is a factor.

Can the remainder theorem be used for $x^2 - 1$?

Not in the simple form $R = f(a)$. The theorem in this article applies to divisors of the form $x - a$. For a higher-degree divisor, use polynomial long division or another appropriate polynomial-division method.

Why might a very small nonzero result show as $0$?

Numerical calculations can produce tiny values that are effectively zero for display purposes. This calculator treats values with absolute size below $10^{-10}$ as zero when displaying the result and checking whether the divisor is a factor.

Can I use fractions in the polynomial?

Use decimal coefficients if the input field does not accept slash-style fractions. For example, use 0.5x^2 instead of 1/2x^2. If you need an exact fractional result, keep an exact written solution or use a symbolic algebra method.

Sources and References

Books

Jay Abramson et al. College Algebra 2e. OpenStax, 2021. Chapter 5, Section 5.5, “Zeros of Polynomial Functions,” especially the remainder theorem, factor theorem, and synthetic-division examples. OpenStax College Algebra 2e
Carl Stitz and Jeff Zeager. College Algebra. Version $\lfloor \pi \rfloor$, Corrected Edition, 2013. Section 3.2, “The Factor Theorem and the Remainder Theorem,” especially synthetic division by $x - c$, quotient and remainder interpretation, and factor checks. Stitz-Zeager College Algebra PDF

Online Educational Sources

Eric W. Weisstein. “Horner's Rule.” MathWorld—A Wolfram Resource, accessed July 4, 2026. MathWorld: Horner's Rule