|Class 9 Maths Polynomials||Remainder Theorem|
It is little tedious to divide 2 polynomials using the process described in previous section, So we use Remainder theorem.
Remainder theorem: Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a).
Proof: Let p(x) be any polynomial with degree greater than or equal to 1. Suppose that when p(x) is divided by x – a, the quotient is q(x) and the remainder is r(x), i.e., p(x) = (x – a) q(x) + r(x) -- (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. Thus we can re-write eq (i) as p(x) = (x – a) q(x) + r –(ii)
In particular, if x = a, then eq (ii) becomes p(a) = (a – a) q(a) + r = r,
Question: Find the remainder when p(x) = x4 + x3 – 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(1) .
So, p(1) = (1)4 + (1)3 – 2(1)2 + 1 + 1 = 2