Remainder Theorem
Remainder Theorem
The assertion that P(c) is the remainder when polynomial P(x) is divided by x – c.
See also
Synthetic division, polynomial long division, polynomial facts
Key Formula
P(x)=(x−c)Q(x)+P(c)
Where:
- P(x) = The polynomial being divided (the dividend)
- x−c = The linear divisor, where c is a constant
- Q(x) = The quotient polynomial resulting from the division
- P(c) = The remainder, found by evaluating P at x = c
Worked Example
Problem: Find the remainder when P(x) = 2x³ − 3x² + 4x − 5 is divided by (x − 2).
Step 1: Identify the value of c from the divisor (x − c). Here the divisor is (x − 2), so c = 2.
c=2
Step 2: By the Remainder Theorem, the remainder equals P(c) = P(2). Substitute x = 2 into the polynomial.
P(2)=2(2)3−3(2)2+4(2)−5
Step 3: Compute each term: 2(8) = 16, −3(4) = −12, 4(2) = 8, and the constant is −5.
P(2)=16−12+8−5
Step 4: Add the results to find the remainder.
P(2)=7
Answer: The remainder when 2x³ − 3x² + 4x − 5 is divided by (x − 2) is 7.
Another Example
This example shows how the Remainder Theorem connects to the Factor Theorem: if P(c) = 0, then (x − c) is a factor. Here the divisor involves (x + 3), so students must correctly identify c = −3, a common source of sign errors.
Problem: Determine whether (x + 3) is a factor of P(x) = x³ + 2x² − 5x + 6.
Step 1: Rewrite the divisor in the form (x − c). Since x + 3 = x − (−3), we have c = −3.
c=−3
Step 2: Evaluate P(−3) to find the remainder.
P(−3)=(−3)3+2(−3)2−5(−3)+6
Step 3: Compute each term: (−3)³ = −27, 2(9) = 18, −5(−3) = 15, and the constant is 6.
P(−3)=−27+18+15+6
Step 4: Add the results.
P(−3)=12
Step 5: Since the remainder is 12, not 0, (x + 3) is NOT a factor of P(x).
P(−3)=0
Answer: (x + 3) is not a factor of x³ + 2x² − 5x + 6 because the remainder is 12, not 0.
Frequently Asked Questions
What is the difference between the Remainder Theorem and the Factor Theorem?
The Factor Theorem is a special case of the Remainder Theorem. The Remainder Theorem says the remainder of P(x) ÷ (x − c) is P(c). The Factor Theorem adds: if that remainder P(c) equals 0, then (x − c) is a factor of P(x). So the Factor Theorem answers the specific question "Is (x − c) a factor?" while the Remainder Theorem gives you the actual remainder value.
Does the Remainder Theorem work for any divisor?
The standard Remainder Theorem applies only when the divisor is a linear expression of the form (x − c). It does not directly apply to divisors like (x² + 1) or (2x − 3). However, for a divisor like (2x − 3), you can factor out the 2 and adapt the approach, since the remainder when dividing by (2x − 3) equals P(3/2).
When should you use the Remainder Theorem instead of polynomial long division?
Use the Remainder Theorem when you only need the remainder and do not need the quotient. It is much faster because you simply plug a number into the polynomial. If you need the full quotient polynomial as well, use synthetic division or polynomial long division instead.
Remainder Theorem vs. Factor Theorem
| Remainder Theorem | Factor Theorem | |
|---|---|---|
| Definition | The remainder when P(x) is divided by (x − c) equals P(c) | (x − c) is a factor of P(x) if and only if P(c) = 0 |
| What it tells you | The exact numerical value of the remainder | Whether a given linear expression is a factor (yes or no) |
| Formula | Remainder = P(c) | P(c) = 0 ⟹ (x − c) is a factor |
| Relationship | The general theorem | A special case of the Remainder Theorem (when remainder = 0) |
| When to use | To quickly find the remainder without full division | To test whether a specific value is a root / factor |
Why It Matters
The Remainder Theorem appears frequently in Algebra 2 and Precalculus when you work with polynomial division, factor polynomials, or find roots. It saves significant time on tests: instead of performing synthetic or long division, you evaluate P(c) with a single substitution. It also forms the logical foundation for the Factor Theorem and the Rational Root Theorem, both of which are essential tools for factoring higher-degree polynomials.
Common Mistakes
Mistake: Using the wrong sign for c when the divisor is (x + c). For example, if dividing by (x + 3), students substitute c = 3 instead of c = −3.
Correction: Always rewrite the divisor in the form (x − c) first. Since x + 3 = x − (−3), the correct value is c = −3. Then evaluate P(−3).
Mistake: Confusing the remainder with the quotient. Students sometimes perform synthetic division correctly but then read off the quotient coefficients instead of the final number, which is the remainder.
Correction: In synthetic division, the remainder is always the last number in the bottom row. When using the Remainder Theorem directly, you skip the quotient entirely — P(c) gives you only the remainder.
Related Terms
- Remainder — The value P(c) that this theorem computes
- Polynomial — The type of expression P(x) this theorem applies to
- Synthetic Division — An efficient division method that also yields the remainder
- Polynomial Long Division — A full division algorithm giving both quotient and remainder
- Polynomial Facts — Collection of key results including this theorem
- Factor Theorem — Special case where remainder equals zero
- Rational Root Theorem — Uses the Remainder Theorem to test candidate roots