Polynomial Remainder Calculator
Find the Remainder of a Polynomial Division
Enter a polynomial P(x) and a divisor x − a (example: P(x) = x³ − 6x² + 11x − 6, x − 2) and click Calculate Remainder.
x − a.
Polynomial Remainder – Complete Explanation
The Remainder Theorem is one of the most useful shortcuts in algebra. It states that if a polynomial P(x) is divided by x − a, then the remainder is P(a). This provides a fast way to find remainders without performing long division.
The Remainder Theorem
If a polynomial P(x) is divided by (x − a), the remainder is P(a).
If P(a) = 0, then (x − a) is a factor of P(x).
1. Steps to Use the Remainder Theorem
- Write the polynomial P(x) in standard form (descending powers)
- Identify 'a' from the divisor x − a (note: if divisor is x + a, then a = −a)
- Evaluate P(a) by substituting x = a into the polynomial
- The result is the remainder
- If remainder = 0, then (x − a) is a factor of P(x)
2. Step-by-Step Examples
Example 1: Remainder = 0 (Factor)
P(x) = x³ − 6x² + 11x − 6 Divisor: x − 2
Step 1: Identify a from x − 2 → a = 2
Step 2: Substitute x = 2 into P(x):
P(2) = 2³ − 6(2)² + 11(2) − 6
= 8 − 6(4) + 22 − 6
= 8 − 24 + 22 − 6
= 0
Remainder: 0
Since the remainder is 0, (x − 2) is a factor of P(x).
Example 2: Non-Zero Remainder
P(x) = x³ − 3x + 2 Divisor: x − 2
Step 1: a = 2
Step 2: P(2) = 8 − 6 + 2 = 4
Remainder: 4
Example 3: Divisor x + a (Negative a)
P(x) = 2x³ − x² + 3x − 5 Divisor: x + 1
Step 1: Rewrite x + 1 as x − (−1) → a = −1
Step 2: Substitute x = −1:
P(−1) = 2(−1)³ − (−1)² + 3(−1) − 5
= 2(−1) − (1) − 3 − 5
= −2 − 1 − 3 − 5
= −11
Remainder: −11
Example 4: Higher Degree Polynomial
P(x) = x⁴ − 2x² + x − 1 Divisor: x − 2
Step 1: a = 2
Step 2: P(2) = 16 − 8 + 2 − 1 = 9
Remainder: 9
3. Synthetic Division (Alternative Method)
Synthetic division is a shortcut method for dividing a polynomial by x − a. The last number in the synthetic division process is the remainder.
Example using Synthetic Division:
Divide x³ − 6x² + 11x − 6 by x − 2
Coefficients: 1 -6 11 -6
Bring down 1, multiply by 2, add, repeat:
1 -6 11 -6
2 | 2 -8 6
----------------
1 -4 3 0 ← Remainder
The remainder is 0, confirming our Remainder Theorem result.
4. Why the Remainder Theorem Works
When we divide P(x) by (x − a), we can write:
P(x) = (x − a) × Q(x) + R
Where Q(x) is the quotient and R is the remainder (a constant). If we substitute x = a:
P(a) = (a − a) × Q(a) + R = 0 × Q(a) + R = R
Therefore, P(a) = R, proving the theorem.
5. Common Mistakes to Avoid
- Sign errors: For divisor x + a, remember a = −a, not +a
- Substitution errors: Carefully evaluate powers and signs
- Missing terms: Include terms with coefficient 0 (e.g., x² + 3 has no x term, so coefficient of x is 0)
- Arithmetic mistakes: Double-check calculations, especially with negatives
- Confusing factor vs. remainder: Remainder = 0 means (x − a) is a factor
6. Connection to the Factor Theorem
The Factor Theorem is a special case of the Remainder Theorem:
The Factor Theorem
(x − a) is a factor of P(x) if and only if P(a) = 0.
This means finding a root (a value that makes P(x) = 0) gives us a factor.
7. Real-World Applications
The Remainder Theorem and polynomial division are used in:
- Checking factors quickly without full division
- Polynomial factorization and simplification
- Root finding for equations
- Engineering computations (control systems, signal processing)
- Computer algorithms (polynomial evaluation, coding theory)
- Cryptography (polynomial-based encryption)
- Physics (modeling with polynomial functions)
8. Practice Problems with Solutions
Problem 1: Find remainder when x² − 5x + 6 is divided by x − 3
Solution: P(3) = 9 − 15 + 6 = 0 → Remainder = 0 (x − 3 is a factor)
Check: x² − 5x + 6 = (x − 2)(x − 3) ✓
Problem 2: Find remainder when 2x³ − x² + 3x − 5 is divided by x + 1
Solution: P(−1) = −2 − 1 − 3 − 5 = −11 → Remainder = −11
Check: Using synthetic division confirms remainder −11
Problem 3: Find remainder when x⁴ − 2x² + x − 1 is divided by x − 2
Solution: P(2) = 16 − 8 + 2 − 1 = 9 → Remainder = 9
Problem 4: Find remainder when x³ + 8 is divided by x + 2
Solution: P(−2) = −8 + 8 = 0 → Remainder = 0 (x + 2 is a factor)
Check: x³ + 8 = (x + 2)(x² − 2x + 4) ✓
Problem 5: Find remainder when 3x⁴ − 2x³ + x² − x + 4 is divided by x − 1
Solution: P(1) = 3 − 2 + 1 − 1 + 4 = 5 → Remainder = 5
9. Advanced Applications
Polynomial Evaluation
The Remainder Theorem gives us an efficient way to evaluate polynomials at specific points. Instead of substituting directly (which can be tedious), we can use synthetic division.
Finding Roots and Factors
If we suspect a root r, we can quickly check if P(r) = 0. If it is, we've found a factor.
Partial Fraction Decomposition
In calculus, finding remainders helps with integration of rational functions.
10. Tips for Mastering the Remainder Theorem
- Always write polynomials in standard form (descending powers)
- Include all terms (use 0 for missing powers)
- Double-check the sign of 'a' when divisor is x + a
- Practice with different degrees (2nd, 3rd, 4th degree)
- Verify results using synthetic division or long division
- Remember: remainder 0 means (x − a) is a factor
- Use the theorem to check factorization quickly
11. Connection to Calculus
The Remainder Theorem is related to Taylor polynomials and approximations. The remainder when dividing by (x − a)ⁿ gives information about derivatives at a.
12. Final Thoughts
The polynomial remainder theorem is a powerful tool that saves time and simplifies calculations. It's elegant, efficient, and forms the foundation for more advanced topics in algebra and calculus.
Use this calculator to practice and verify your work, but always work through problems manually to build confidence and understanding. With consistent practice, you'll be able to find remainders instantly and use this theorem to factor polynomials and solve equations efficiently.