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.
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.
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 \(+1\) \(+1\).
So that, the remainder of the polynomial x4+x32x2+x+1 is \(2\).