The Remainder Theorem
Simple Explanation
The Remainder Theorem says that dividing a polynomial P(x) by (x − a) always leaves a remainder equal to P(a) — so you can find the remainder with a single substitution, without doing any division at all.
Why Do We Need It?
This is a genuine shortcut: it replaces an entire division process with one evaluation, saving significant work whenever only the remainder (not the full quotient) is needed.
Formula
The Remainder Theorem
If P(x) is divided by (x − a), the remainder equals P(a)
Dividing a polynomial by (x − a) always leaves a remainder equal to the polynomial evaluated at x = a — found by a single substitution, without doing the division at all.
- P(x)
- — the polynomial being divided
- a
- — the constant in the linear divisor (x − a)
- P(a)
- — the polynomial evaluated at x = a — this equals the remainder
When to use it: Whenever you need the remainder of a polynomial divided by a linear factor (x − a), without carrying out the full division.
Worked Example
Find a remainder using substitution
Find the remainder when P(x) = x³ − 2x² + 4x − 1 is divided by (x − 2), without dividing.
Why Does This Work?
From the division identity, P(x) = (x−a)Q(x) + R, where R is a constant (since the degree of R must be less than the degree of the linear divisor, degree 1, meaning R has degree 0). Substituting x = a makes the (x−a)Q(x) term vanish entirely, leaving P(a) = R exactly.
Real-Life Example
Quick error-checking in engineering calculations
An engineer wants to quickly verify whether a specific measurement value satisfies a polynomial design constraint, without fully dividing the formula.
A single substitution (the remainder theorem) tells them immediately whether that value gives an exact fit (remainder 0) or how far off it is.
Practice
Find the remainder when P(x) = 2x³ + x − 5 is divided by (x + 1). (Note: x+1 = x−(−1).)
MediumCommon mistake
Applying the remainder theorem to a non-linear divisor (like x² − 4) — it only works directly for linear divisors of the exact form (x − a).
Quick Review
- Dividing P(x) by (x − a) leaves a remainder of exactly P(a).
- This replaces the whole division process with a single substitution.
- Only applies to linear divisors of the form (x − a).