**Remainder Theorem (शेषफल प्रमेय)**

Let p(x) be any polynomial of degree greater than or equal to 1 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. Suppose that when p(x) is divided by x – a, then quotient is q(x) and remainder is r(x). i.e.

Since the degree of (x – a) is 1 then 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 for every value of x, r(x) = r.

therefore, p(x) = (x – a)q(x) + r

If x = a, then the equation will give us:

p(a) = (a – a)q(a) + r = 0 + r

p(a) = 0

Which proves the theorem.

### Polynomial Remainder Theorem Examples With Answers

**Example 1:-**

**Explanation:**

So the remainder will be 4.

**Example 2:-**

**Explanation:**

p(-1) = 1+3+2+4-1

So the remainder will be 9.

**Example 3:-**

**Check whether x – 2 is a factor of x³+x²-2x-8**

**जांच कीजिए कि **x – 2,** x³+x²-2x-8 का एक गुणानखण्ड है या नहीं?**

**Explanation:**

If x – 2 is a factor of x³+x²-2x-8 then when we will divide x³+x²-2x-8 by x – 2 then remainder must be zero.

So by remainder theorem:

p(x) = x³+x²-2x-8 and the zero of x – 2 is 2.

So by remainder theorem we can say that the remainder will be p(2).

p(2) = 2³+2²-2(2)-8

= 8+4-4-8

= 0

So there remainder is zero that means x – 2 is a factor of x³+x²-2x-8

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