An expression of the form \(\mathbf{ax^{n} + bx^{n-1} + cx^{n-2} + ….. + kx + 1}\)

where each variable has a constant accompanying it as its coefficient is called a polynomial of degree ‘n’ in variable x. Each variable separated with an addition or subtraction symbol in the expression is better known as

**term**. Degree is defined as the maximum power of the variable of a polynomial. For example a linear polynomial of the form *ax+b *is called a polynomial of degree 1. Similarly, quadratic polynomials and cubic polynomials have a degree of 2 and 3 respectively.

A polynomial with only one term is known as a monomial. A monomial containing only constant term is said to be a polynomial of zero degree. Is zero a polynomial? Yes, if in a polynomial the value of all the constants equals to zero that is, a=b=c=………………=k=l=0, we call such a polynomial as zero polynomial. A polynomial can account to null value even if the values of the constants are greater than zero. In such cases we look for the value of variables which set the value of entire polynomial to zero. These values of a variable are known as the roots of polynomials.

## Calculation of Roots of Polynomials

Let us take an example of polynomial P(x) of degree 1,

**P(x) = 5x + 1**

According to the definition of roots of polynomials, ‘a’ is the root of a polynomial P(x), if

P(a) = 0.

Thus, in order to determine the roots of polynomial P(x), we have to find the value of x for which P(x) = 0. Now,

5x + 1 = 0

- x= -1/5

Hence, ‘-1/5’ is the root of polynomial P(x).

## Roots of Polynomials: Solved Examples

**Example: Check whether -2 is a root of polynomial 3x ^{3 }+ 5x^{2 }+ 6x + 4.**

**Solution: **According to the question,

P(x) = 3x^{3}+ 5x^{2}+ 6x+ 4

At x = -2,

\(P(-2) = 3(-2)^{3}+ 5 (-2)^{2} + 6(-2) + 4\)

\(\Rightarrow P(-2) = -24 + 20 – 12 + 4 = -12\)

As, P(-2) , hence -2 is not a root of polynomial 3x^{3}+ 5x^{2}+ 6x+ 4.

**Examples: Find the roots of the polynomial \(x^{2} + 2x -15\)**

**Solution: **Given \(x^{2} + 2x -15\)

Breaking the terms, we have,

\(x^{2} + 5x – 3x -15\)

\( = x(x + 5) – 3(x + 5)\)

\(= (x – 3) (x + 5)\)

\(\Rightarrow x = 3 \;\; or \;\; x = -5\)

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