The Remainder Theorem says that when a polynomial f(x) is divided by x − a, the remainder is f(a). You find it by substituting a into the polynomial, with no long division needed. Its partner, the Factor Theorem, says that x − a is a factor exactly when f(a) = 0. This guide gives the formula, a short proof, and worked examples for finding remainders, testing factors, finding unknown coefficients and factoring polynomials.
In this guide
What Is the Remainder Theorem?
The Remainder Theorem says: if a polynomial f(x) is divided by x − a, the remainder is f(a). Instead of dividing, you substitute the number a for x and calculate.
| Part | Meaning | Example: f(x) = x² + 3x + 2 ÷ (x − 1) |
|---|---|---|
| Divisor | x − a | x − 1, so a = 1 |
| Formula | Remainder = f(a) | f(1) |
| Calculation | Substitute a into f(x) | 1 + 3 + 2 |
| Remainder | The result | 6 |
Check it with division: (x² + 3x + 2) ÷ (x − 1) gives x + 4 with remainder 6, because (x − 1)(x + 4) + 6 = x² + 3x − 4 + 6 = x² + 3x + 2. ✓ The method of dividing polynomials is covered in its own guide.
In short: Divide f(x) by x − a and the remainder is f(a). To find a, set the divisor equal to 0: for x − 1 use 1, for x + 2 use −2.
Why the Remainder Theorem Works (Proof)
Division always gives f(x) = (x − a) × q(x) + r, where q(x) is the quotient and r is the remainder. Because the divisor has degree 1, the remainder r is just a number.
Now substitute x = a. The bracket becomes (a − a) = 0, so the whole product disappears:
f(a) = 0 × q(a) + r = r
So the remainder r is exactly f(a). For numbers it is the same idea: 17 = 5 × 3 + 2, so 17 minus the multiple of 5 leaves the remainder 2.
How to Find the Remainder Using the Theorem
- Set the divisor equal to 0 and solve to find a. For x + 2, a = −2. For 2x − 1, a = 1/2.
- Substitute a into f(x). Use brackets, especially for negative values.
- Calculate. The answer is the remainder.
In short: Set the divisor to 0 to find a, substitute a into the polynomial, and calculate. The result is the remainder.
Remainder Theorem Examples
Example 1: divisor with a negative a. Find the remainder when f(x) = x³ + 4x² − x + 5 is divided by x + 2.
Set x + 2 = 0, so a = −2. Then f(−2) = (−2)³ + 4(−2)² − (−2) + 5 = −8 + 16 + 2 + 5 = 15. Synthetic division with a = −2 on the coefficients 1, 4, −1, 5 gives the bottom row 1, 2, −5, 15, which confirms it.
Example 2: divisor with a leading coefficient. Find the remainder when f(x) = 4x³ − 3x + 5 is divided by 2x − 1.
Set 2x − 1 = 0, so a = 1/2. Then f(1/2) = 4(1/8) − 3(1/2) + 5 = 0.5 − 1.5 + 5 = 4.
Example 3: find an unknown coefficient. The polynomial f(x) = x³ + kx² − 2x + 4 leaves remainder 6 when divided by x − 2. Find k.
By the theorem f(2) = 6. So 8 + 4k − 4 + 4 = 6, which gives 8 + 4k = 6 and k = −1/2. Check: f(2) = 8 − 2 − 4 + 4 = 6. ✓
Example 4: a remainder of zero. Find the remainder when f(x) = x³ − 2x² − 5x + 6 is divided by x − 3. Here f(3) = 27 − 18 − 15 + 6 = 0. A zero remainder means something special, as the next section shows.
In short: Always find a first. For a divisor like 2x − 1, a is a fraction (1/2). To find an unknown coefficient, set f(a) equal to the given remainder and solve.
The Factor Theorem
The Factor Theorem is the special case where the remainder is 0: x − a is a factor of f(x) if and only if f(a) = 0. In other words, a is a root (zero) of the polynomial.
Example 5. Is x − 3 a factor of x³ − 2x² − 5x + 6? Since f(3) = 0 (Example 4), yes.
Example 6. Is x + 1 a factor of x³ + 2x² + x + 3? Here a = −1 and f(−1) = −1 + 2 − 1 + 3 = 3, which is not 0, so no. The remainder is 3.
In short: If f(a) = 0, then x − a is a factor. If f(a) is not 0, it is the remainder and x − a is not a factor.
Factoring Polynomials With the Factor Theorem
This is the most useful application. For a cubic or higher polynomial, test small values until f(a) = 0. Then divide by x − a to get a smaller polynomial, and factor that.
Example 7: factor f(x) = x³ − 6x² + 11x − 6.
- Try a = 1: f(1) = 1 − 6 + 11 − 6 = 0, so x − 1 is a factor.
- Divide by x − 1 with synthetic division: the coefficients 1, −6, 11, −6 give the bottom row 1, −5, 6, 0.
- The quotient is x² − 5x + 6 = (x − 2)(x − 3).
- So f(x) = (x − 1)(x − 2)(x − 3), and the roots are 1, 2 and 3.
Example 8: solve x³ − 7x + 6 = 0. f(1) = 1 − 7 + 6 = 0, so x − 1 is a factor. Dividing with the coefficients 1, 0, −7, 6 gives 1, 1, −6, 0, so the quotient is x² + x − 6 = (x + 3)(x − 2). The solutions are x = 1, x = 2 and x = −3. You can confirm them with the solve for x calculator.
In short: Test values until f(a) = 0, divide by x − a, then factor what is left. Each root found removes one degree.
Which Values of a Should You Test?
You do not have to guess. For a polynomial with whole-number coefficients, any rational root has the form p/q, where p divides the constant term and q divides the leading coefficient. This is the Rational Zero Theorem.
Example 9: f(x) = 2x³ − 3x² − 3x + 2. The constant term 2 has factors ±1, ±2, and the leading coefficient 2 has factors ±1, ±2. The candidates are ±1, ±2 and ±1/2.
| Test value a | f(a) | Factor? |
|---|---|---|
| 1 | 2 − 3 − 3 + 2 = −2 | No |
| 2 | 16 − 12 − 6 + 2 = 0 | Yes, x − 2 |
| −1 | −2 − 3 + 3 + 2 = 0 | Yes, x + 1 |
| 1/2 | 0.25 − 0.75 − 1.5 + 2 = 0 | Yes, 2x − 1 |
So f(x) = (x − 2)(x + 1)(2x − 1). The roots are 2, −1 and 1/2.
Remainder Theorem vs Factor Theorem
| Remainder Theorem | Factor Theorem | |
|---|---|---|
| Statement | Dividing f(x) by x − a leaves remainder f(a) | x − a is a factor if and only if f(a) = 0 |
| Use it to | Find a remainder without dividing | Test factors and find roots |
| Result | Any number | Always 0 |
| Relationship | The general rule | The special case where the remainder is 0 |
Common Mistakes With the Remainder Theorem
- Using the wrong sign for a. For x + 2, a is −2, not 2.
- Forgetting the leading coefficient. For 2x − 1, a = 1/2, not 1.
- Sign errors with negative values. (−2)³ = −8, but (−2)² = 4. Use brackets.
- Thinking the remainder is a polynomial. When you divide by a linear factor, the remainder is a single number.
- Mixing up the two theorems. A non-zero f(a) is a remainder, not a root.
- Using the shortcut for the wrong divisor. The Remainder Theorem in this form needs a linear divisor, such as x − a or px − q.
Remainder Theorem Practice Problems (With Answers)
1. Find the remainder when x² − 5x + 7 is divided by x − 2.
1. f(2) = 4 − 10 + 7 = 1.
2. Find the remainder when x³ − 2x² + x − 3 is divided by x − 3.
9. f(3) = 27 − 18 + 3 − 3 = 9.
3. Find the remainder when 2x³ + x² − 4 is divided by x + 1.
−5. f(−1) = −2 + 1 − 4 = −5.
4. Is x + 3 a factor of x³ + 3x² − 4x − 12?
Yes. f(−3) = −27 + 27 + 12 − 12 = 0.
5. Is x − 2 a factor of x³ − 4x² + x + 6?
Yes. f(2) = 8 − 16 + 2 + 6 = 0.
6. Find k so that x − 1 is a factor of x³ + kx² − x + 2.
k = −2. f(1) = 1 + k − 1 + 2 = k + 2 = 0.
7. Find the remainder when x⁴ − 3x² + 2 is divided by x + 2.
6. f(−2) = 16 − 12 + 2 = 6.
8. Factor x³ − x² − 4x + 4.
(x − 1)(x − 2)(x + 2). f(1) = 0, then dividing by x − 1 leaves x² − 4.
Remainder Theorem: Frequently Asked Questions
What is the remainder theorem?
The remainder theorem states that when a polynomial f(x) is divided by x − a, the remainder is f(a). You can find the remainder by substituting a into the polynomial instead of doing long division.
What is the remainder theorem formula?
Remainder = f(a), where the divisor is x − a. To find a, set the divisor equal to 0. For x + 2, a = −2, and for 2x − 1, a = 1/2.
What is the difference between the remainder theorem and the factor theorem?
The remainder theorem gives the remainder when you divide by x − a, which is f(a). The factor theorem is the special case: x − a is a factor of f(x) if and only if the remainder f(a) is 0.
How do you use the remainder theorem to factor a polynomial?
Test values of a until f(a) = 0. Then x − a is a factor. Divide by x − a, usually with synthetic division, and factor the quotient. Repeat until the polynomial is fully factored.
Does the remainder theorem work for any divisor?
The form f(a) works for linear divisors x − a, or px − q using a = q/p. For divisors of higher degree, such as x² + 1, you need to use long division.
Why is the remainder a constant?
A remainder always has a lower degree than the divisor. When the divisor is linear (degree 1), the remainder must have degree 0, which means it is a single number.
Sources and Further Reading
- OpenStax, College Algebra 2e, Section 5.5: Zeros of Polynomial Functions, Rice University.
- Math is Fun, Remainder Theorem and Factor Theorem.
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