Class 9 Maths Polynomials | Remainder Theorem |

**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) = x^{4 }+ x^{3} – 2x^{2} + 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

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