Remainder Theorem
Let p(x) be any polynomial of degree n>0 ,and a any real number. If p(x) is divided by ( x-a), then the remainder is p(a).

The Remainder Theorem can be proved as follows.

Proof
Let us suppose when p(x) is divided by ( x-a), the quotient is q(x) and remainder is r(x).

So we have,
p(x)=(x-a) q(x)+r(x), where r(x)=0 or degree of r(x)< degree of x-a.
Since degree of ( x-a) is1, either r(x)=0 or degree of r(x)=0
So r(x) must be a constant,say r.

Thus for all values of x,
p(x)=(x-a) q(x)+r ........(1), where r is a constant.

In particular, when x=a,
p(a)=0.q(x)+r
=0+r
=r
Hence the theorem.

Factor Theorem
Let p(x) be a polynomial of degree n>0. If p(a)=0 for a real number a, then (x-a) is a factor of
p(x). Conversely, if (x-a) is factor of p(x), then p(a)=0.

Proof

First part:
Let p(a)=0
Then by remander theorem, r=0
So equation (1) becomes
p(x)=(x-a) q(x)
==>(x-a) is a factor of p(x).

Second Part:
By remainder theorem,

p(x)=(x-a) q(x)+r
ie. p(x)=(x-a) q(x)+p(a)
Since (x-a) is a factor, p(a) must be zero.
This proves the theorem.

Example
Find the remainder when p(x)= x^2 +3x+1 is divided by x+1.
Determine whether( x-2) is a factor of p(x) or not.

Solution
x+1= x-(-1)
[Always check whether the divisor is in the form of (x-a)or not. Otherwise rewrite that in the form of(x-a)]

So, here a=-1
There fore by remainder theorem, the required remainder is
p(a)= p(-1)
=(-1)^2+3(-1)+1
=1-3+1
=-1

We know, by factor theorem,

if (x-2) is a factor of p(x), then p(2) must be zero.

Here p(2)=2^2+3(2)+1
=4+6+1
=11 which is not equal to zero.
So (x-2) is not a factor of p(x).