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)=2x3-5x+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)=2x3-5x+3 is divided by
P(1)=2·13-5·1+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.