# Remainder Theorem - Statement, Proof & Examples

## TL;DR

The remainder theorem says that when a polynomial p(x) is divided by (x−a), the remainder is just p(a) — so you find a remainder by substituting, not by dividing. This article states and proves the theorem, shows how it connects to the factor theorem, works six examples, and names the sign and substitution mistakes that cost the most marks.

## What Is The Remainder Theorem?

**The remainder theorem states that when a polynomial p(x) is divided by a linear divisor (x−a), the remainder equals p(a) — the value of the polynomial at x=a.** Because the divisor is linear (degree 1), the remainder is always a constant.

In symbols, dividing p(x) by (x−a) gives a quotient q(x) and a remainder r:

p(x)=(x−a)q(x)+r

and the theorem tells you that r=p(a).

### Variable Glossary:

| Symbol       | Meaning                                       |
|--------------|-----------------------------------------------|
| p(x)        | the polynomial being divided (the dividend)  |
| (x−a)       | the linear divisor                            |
| q(x)        | the quotient                                  |
| r            | the remainder, a constant                     |
| a            | the number that makes the divisor zero       |

A useful extension: when the divisor is written as (ax+b) rather than (x−a), the remainder is p(−b/a) — you substitute the value that makes the divisor zero.

### The link to synthetic division

There is a second way to find the same remainder: **synthetic division**, a compact shorthand for dividing by (x−a) using only the coefficients. You write the dividend's coefficients in a row, bring down the leading one, multiply by a, add to the next coefficient, and repeat. The last number you produce is the remainder — and it equals p(a), exactly what the remainder theorem promises. Synthetic division is the faster route when you also want the quotient; pure substitution is faster when you want the remainder alone. Both rest on the same identity p(x) = (x−a)q(x) + r.

## How Do You Prove The Remainder Theorem?

The proof is short and rests on the division algorithm. Any polynomial p(x) divided by (x−a) can be written as:

p(x)=(x−a)q(x)+r

where r is a constant because the divisor has degree 1, so the remainder has degree 0.

This identity holds for every value of x, so substitute x=a:

p(a)=(a−a)q(a)+r

The factor (a−a) is 0, which kills the first term:

p(a)=0⋅q(a)+r=r

So r=p(a). The remainder equals the polynomial evaluated at a.

## Examples of Remainder Theorem

### Example 1

**Find the remainder when p(x)=x²+5x+6 is divided by (x−2).**

Substitute x=2:

p(2)=2²+5(2)+6

=4+10+6=20

**Final answer:** the remainder is 20.

### Example 2

**Find the remainder when p(x)=x³−4x+1 is divided by (x+2).**

Substitute x=−2:

p(−2)=(-2)³−4(−2)+1

=−8+8+1=1

**Final answer:** the remainder is 1.

### Example 3

**Find the remainder when p(x)=2x³+x²−5x+3 is divided by (x−1).**

Substitute x=1:

p(1)=2(1)³+(1)²−5(1)+3

=2+1−5+3=1

**Final answer:** the remainder is 1.

### Example 4

**Find the remainder when p(x)=x³+3x²−2x+4 is divided by (2x−1).**

Substitute x=1/2:

p(1/2)=(1/2)³+3(1/2)²−2(1/2)+4

=1/8 + 3(1/4) - 1 + 4 = 31/8

**Final answer:** the remainder is 31/8.

### Example 5

**Use the remainder theorem to check whether (x−3) is a factor of p(x)=x³−7x+6.**

Substitute x=3:

p(3)=3³−7(3)+6

=27−21+6=12

**Final answer:** (x−3) is not a factor, since p(3)=12≠0.

### Example 6

**Find the value of k if the remainder is 5 when p(x)=x²+kx+2 is divided by (x−3).**

p(3)=3²+k(3)+2=9+3k+2=11+3k

Set equal to 5:

11+3k=5

3k=−6 
k=−2

**Final answer:** k=−2.

## Key Takeaways

- The remainder theorem says the remainder of p(x) divided by (x−a) equals p(a), found by substitution, not division.
- The proof follows from the division identity p(x)=(x−a)q(x)+r evaluated at x=a.
- The factor theorem is the special case: p(a)=0 means (x−a) is a factor.
- The theorem works only for linear divisors; substitute the zero of the divisor, signs included.
- It is the engine behind finding roots of higher-degree polynomials without full long division.