Remainder theorem

**Remainder theorem.**

If p(x) is divides by (x-c) the remainder is a constant and and is equal to p(c)

Supose that we divide the polynomial P(x)=2x^{3}-5x+3 by x=1.

Then according to the remainder theorem, the remainder in this case should be the number p(1). Let's check:

The remainder is 0 and P(1) = 0.

As the calculations show, the remainder is indeed equal to p(1)

Use the remainder theorem to find the remainder when P(x)=2x^{3}-5x+3 is divided by

x-1.

P(1)=2·1^{3}-5·1+3=2-5+3=0. Thus the remainder is 0.

**Proof of the remainder theorem**

To prove the remainder theorem we must show that when the polynomial P(x) is divided by x-a, the remainder is P(a).

Now, according to the division algorithm, we can write P(x)=(x-a)·q(x) + r(x) (1) for unique polynomials q(x) and r(x). In this identity, either r(x) is the zero polynomial or the degree of r(x) is less than that of x-a. Since the degree of x-a is 1, the degree of r(x) must be zero. Thus in either case, the remainder r(x) is a constant. Denoting this constant by r, we can rewrite equation (1) as:P(x)=(x-a)·C(x) + r.

If we set x=a in this identity we obtain P(a)=(a-a)·C(a) + r =r .

We have now shown that P(a) = r. But by definition, r is the remainder r(x). Thus P(a) is the remainder.

This proves the remainder theorem.