Theory:

The process described in the previous section is somewhat tedious to divide two polynomials, so we use Remainder theorem.
Remainder theorem:
 
Let p(x) be any degree polynomial greater than or equal to one and let \(a\) be any real number. When a linear polynomial x – a divides p(x), then p(a) will be the Remainder.
 
Proof:
 
Let p(x) be any polynomial that is greater than or equal to degree \(1\).
 
Assume that the quotient is \(q(x)\) when p(x) is divided by x – a and, the remainder is  r(x).
 
Applying the division algorithm, Dividend \(=\) Divisor \(\times\) Quotient \(+\) Remainder
 
p(x) = (x – aq(x) + r(x)  ......................... Equation (i).
 
Since the degree of x – a is \(1\) and the degree of  r(x) is less than the degree of x – a, the degree of r(x)=0.
 
This means that  r(x) is a constant, say r.
 
So we can rewrite Equation (i) as p(x) = (x – aq(x) + r   .................... Equation (ii)
 
In particular, if x = a, then Equation (ii) becomes  p(a) = (a – aq(a) + r  =  r.
 
Thus, \(p(a) = r\).  Therefore, \(p(a)\) is the remainder.
Example:
Find the remainder when p(x) = x 4x 3– 2x2 + x + 1 is divided by x – 1.
 
Solution: Zero of  x – 1 is \(1\), so as per remainder theorem remainder in this case will be \(p(x) =\) \(p(1)\) .
 
Consider the polynomial p(x) =x4+x32x2+x+1.
 
Put x=1.
 
p(1) =(1)4+(1)32(1)2 \(+1\) \(+1\).
 
p(1)=1+12+1+1p(1)=22+1+1p(1)=22+2p(1)=2
 
So that, the remainder of the polynomial x4+x32x2+x+1 is \(2\).