# Remainder Theorem – Polynomials | Class 9 Maths

Polynomial is an algebraic expression consisting of variable and coefficient. Variable is also at times called indeterminate. We can perform any of the operations using polynomials whether it be multiplication, division, subtraction or addition. Examples of polynomial with one variable are x^{2 }+ x – 8, y^{3 }+ y^{2 }– 52, z^{2}+64

The word polynomial was derived from the Greek word ‘**poly**’ meaning ‘many’ and ‘**nominal**’ meaning ‘terms’, so altogether it is said as “**many terms**”. A polynomial can not have infinite terms.

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**Remainder Theorem**

Let **g(x)** be a polynomial of degree 1 or greater than 1 and let b be any real number. If **g(x)** is divided by the linear polynomial x – b, then the remainder is **p(b)**.

**Proof**

Let g(x) be a polynomial with degree 1 or greater than 1. Suppose that when g(x) is divided by (x – b), the quotient is q(x) and the remainder is r(x), i.e.,

g(x) = (x – b) q(x) + r(x) ……(1)

Since the degree of x – b is 1 and that of r(x) is less than the degree of x – b, the degree of r(x) = 0. This means that r(x) is constant. Thus, we can write the equation (1) as

g(x) = (x – b) q(x) + r ……(2)

In particular, if x = b, then equation (2) becomes

p(b) = (b – b) q(b) + r => r

### Sample Problems on Remainder Theorem

**Problem 1: Find the remainder when g(x) = x ^{4} – x^{3} + x^{2} – 2x + 1 is divided by x – 2.**

**Solution:**

Zero of x – 2 is 2, so as per the remainder theorem

Remainder, in this case, will be g(2).

So,

g(2) = (2)

^{4}– (2)^{3}+ (2)^{2}+ 2(2) + 1 = 17

**Problem 2: Find the root of the polynomial x ^{2} – 5x + 4**

**Solution: **

The approach of solving such questions include to choose a number in such a manner, that putting it to use, will yield a zero remainder.

f(x) = x

^{2 }– 5x + 4f(4) = 4

^{2 }– 5(4) + 4f(4) = 20 – 20 = 0

So, (x – 4) must be a factor of x

^{2 }– 5x + 4

**Problem 3: Find the remainder when t ^{3} – 2t^{2 }+ 4t + 5 is divided by t – 1.**

**Solution: **

Since, here it is already given that we need to find the remainder when the given quotient is divided by t – 1. So, accordingly we will put 1 in place of x, to solve and get the remainder.

Here, p(t) = t

^{3}– 2t^{2 }+ 4t + 5, and the zero of t – 1 is 1∴ g(1) = (1)

^{3}– 2(1)^{2}+ 4 + 5 = 8By the Remainder Theorem, 8 is the remainder when t

^{3 }– 2t^{2 }+ 4t + 5 is divided by t – 1

**Problem 4: Find the remainder when x ^{3 }–^{ }x^{2 }+ 2 is divided by x – 2.**

**Solution:**

Here, we will put x = 2 in the given quotient to find the remainder

2^3 – 2^2 + 2

= 8 – 4 + 2

= 6

By the Remainder Theorem, 6 is the remainder when x

^{3}– x^{2 }+ 2 is divided by x – 2.

**Problem 5: By what should, x ^{3} – x^{2} – 4 be divided to give 0 as remainder.**

**Solution:**

We will use Hit & Trial method to find the answer,

Clearly putting x = 2, will give zero as remainder.

So, this will be the answer to yield a zero remainder on division by x – 2.