# Proofs involving the Remainder theorem

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

If $P(x)$ is a polynomial function, then the remainder of $P(x)$ when divided by $x-k$ is equal to $P(k)$.

### Example

Say we wanted to find the remainder of $P(x) = x^3 - 2x^2 - 6x - 7$ divided by $x-4$. We evaluate $P(4)$ 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 $n^{th}$ 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 $P(x)$ after division results in $D(x) \cdot Q(x) + R$ where D is the divisor, Q is the quotient, and R is the remainder.

In this case, the polynomial is divided by $x-k$ where k is a constant. This is the divisor, so $D(x) = x-k$.

Substituting the divisor into the polynomial, we have: $\begin{array}{l} P(x) = D(x) \cdot Q(x) + R \\ P(x) = (x-k) \cdot Q(x) + R \end{array}$

From here, you can easily see that $P(k) = R$: $\begin{array}{l}P(k) = (k-k) \cdot Q(x) + R \\ P(k) = R \end{array}$

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 $P(x) = x^n - y^n$. The variable y is still a variable, but takes the position of k in the remainder theorem.

Evaluating $P(y)$, we get 0. Therefore the remainder of $P(x)$ when divided by $x-y$ is zero, meaning $x-y$ is a factor of $P(x)$.

### Why x^n + y^n is sometimes but not always divisible by x+y Again with the similar polynomial $P(x) = x^n + y^n$, we have x-k as a factor if $P(k)=0$.

The expression x+y is divisible only when $P(-y) = 0$. In other words: $\begin{array}{l} (-y)^n + y^n = 0 \\ (-y)^n = -y^n \end{array}$

If n is even: $(-y)^n = y^n$

However if n is odd: $(-y)^n = -y^n$

Therefore, it’s only when n is odd that $x^n + y^n$ is divisible by $x+y$.

## 2 thoughts on “Proofs involving the Remainder theorem”

1. Anonymous says:

Doubt clearing

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1. Anonymous says:

Woooow

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