The Core Idea

Divide any polynomial f(x) by a linear expression like (x - a), and you get a quotient along with some leftover remainder. That remainder is always a single number, and it turns out to equal f(a) exactly. So instead of dividing at all, you can just substitute a for x and evaluate.

Dividing f(x) by (x - a) gives:
Remainder = f(a)

Why This Actually Works

Any polynomial division can be expressed as f(x) = (x - a) · q(x) + r, where q(x) is the quotient and r is the remainder. Set x equal to a, and the (x - a) factor disappears, since (a - a) = 0 wipes out that entire term. What's left is f(a) = r, which is exactly the claim: the remainder equals f(a).

Example 1: A Remainder of Zero

Suppose you need the remainder when f(x) = x³ - 4x² + x + 6 is divided by (x - 2).

Just evaluate f(2):

f(2) = (2)³ - 4(2)² + (2) + 6
f(2) = 8 - 16 + 2 + 6
f(2) = 0

Remainder = 0

A remainder of zero tells you more than just the answer: (x - 2) divides evenly into f(x), making it a factor. You can verify this directly: x³ - 4x² + x + 6 = (x - 2)(x² - 2x - 3).

Example 2: A Nonzero Remainder

Next, find the remainder when f(x) = 2x³ + 3x² - x + 5 is divided by (x + 1). Since (x + 1) is the same as (x - (-1)), here a = -1.

Evaluate f(-1):

f(-1) = 2(-1)³ + 3(-1)² - (-1) + 5
f(-1) = 2(-1) + 3(1) + 1 + 5
f(-1) = -2 + 3 + 1 + 5
f(-1) = 7

Remainder = 7

How This Ties Into the Factor Theorem

The factor theorem is really just a special case of what you've already seen. It states that (x - a) is a factor of f(x) - meaning it divides in evenly - exactly when f(a) = 0. A remainder of zero is the signal that the division came out clean.

That makes this a fast way to factor higher-degree polynomials: try candidate values, substitute each into the polynomial, and watch for a result of zero. Every zero you hit uncovers another factor.

Where This Comes In Handy

  • Testing possible roots: Check whether a number is a root without factoring the entire polynomial first.
  • Skipping long division: When the remainder is all you need, there's no reason to carry out the full division.
  • Breaking down polynomials: Combined with the factor theorem, it speeds up factoring messy higher-degree expressions.
  • Beyond algebra: The same reasoning shows up in modular arithmetic and gets applied in parts of computer science.

When It Won't Help

This shortcut only applies to division by a linear binomial (x - a). Once the divisor becomes quadratic or higher, the trick stops working - you're back to long division or synthetic division, since the remainder in those cases is no longer simply f(a).