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).


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:

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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

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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

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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

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