The **Remainder theorem**, sometimes called the *Little Bezout theorem* states the following:

If is a polynomial function, then the remainder of when divided by is equal to .

### Example

Say we wanted to find the remainder of divided by . We evaluate which gives 1.

The other way to find the remainder would be by polynomial long division:

This calculation is somewhat wasted if we only want to calculate the remainder, as it takes considerably more work than evaluating a polynomial.

Notice however that if the goal was to obtain the remainder, Synthetic division would be the better method.

To calculate the remainder of a degree polynomial requires *n* multiplications and *n-1* additions, for both the remainder theorem and synthetic division.

In addition, division gives both the quotient and the remainder while evaluating the polynomial gives only the remainder.

### Proof of the Remainder Theorem

Consider that a polynomial after division results in where *D* is the divisor, *Q* is the quotient, and *R* is the remainder.

In this case, the polynomial is divided by where *k* is a constant. This is the divisor, so .

Substituting the divisor into the polynomial, we have:

From here, you can easily see that :

*Q.E.D.* (not sure if this is correct usage or not)

### Why x^n – y^n is divisible by x-y

No idea why my writing tends to slide downwards to the right. I don’t have this problem on paper, only when drawing stuff with the mouse on MS-paint.

Anyways, we start with the polynomial . The variable *y* is still a variable, but takes the position of *k* in the remainder theorem.

Evaluating , we get 0. Therefore the remainder of when divided by is zero, meaning is a factor of .

### Why x^n + y^n is sometimes but not always divisible by x+y

Again with the similar polynomial , we have *x-k* as a factor if .

The expression *x+y* is divisible only when . In other words:

If *n* is even:

However if *n* is odd:

Therefore, it’s only when *n* is odd that is divisible by .