# Roots of Polynomials

**Roots of polynomials** are the solutions for any given polynomial for which we need to find the value of the unknown variable. If we know the roots, we can evaluate the value of polynomial to zero. An expression of the form a_{n}x^{n }+ a_{n-1}x^{n-1 }+ …… + a_{1}x + a_{0}, 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 the **term**. The degree of the polynomial 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 a constant term is said to be a polynomial of zero degrees. 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. Sometimes they are also termed as zeros of polynomials.

## Roots of Polynomials Formula

The polynomials are the expression written in the form of:

a_{n}x^{n}+a_{n-1}x^{n-1}+……+a_{1}x+a_{0}

The formula for the root of linear polynomial such as ax + b is

**x = -b/a**

The general form of a quadratic polynomial is ax^{2 }+ bx + c and if we equate this expression to zero, we get a quadratic equation, i.e. ax^{2 }+ bx + c = 0.

The roots of quadratic equation, whose degree is two, such as ax^{2 }+ bx + c = 0 are evaluated using the formula;

**x = [-b ± √(b ^{2 }– 4ac)]/2a**

The formulas for higher degree polynomials are a bit complicated.

### Roots of three-degree polynomial

To find the roots of the three-degree polynomial we need to factorise the given polynomial equation first so that we get a linear and quadratic equation. Then, we can easily determine the zeros of the three-degree polynomial. Let us understand with the help of an example.

Example: 2x^{3 }− x^{2 }− 7x + 2

Divide the given polynomial by x – 2 since it is one of the factors.

2x^{3 }− x^{2 }− 7x + 2 = (x – 2) (2x^{2} + 3x – 1)

Now we can get the roots of the above polynomial since we have got one linear equation and one quadratic equation for which we know the formula.

**Also, read:**

### Finding Roots of Polynomials

Let us take an example of the polynomial p(x) of degree 1 as given below:

**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 the polynomial p(x).

### Questions and Solutions

**Example 1: Check whether -2 is a root of polynomial 3x ^{3 }+ 5x^{2 }+ 6x + 4.**

**Solution: **Let the given polynomial be,

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

Substituting x = -2,

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

p(-2) = -24 + 20 – 12 + 4 = -12

Here, p(-2) ≠ 0

Therefore, -2 is not a root of the polynomial 3x^{3 }+ 5x^{2 }+ 6x + 4.

**Example 2: Find the roots of the polynomial x ^{2} + 2x – 15**

**Solution: **Given x^{2} + 2x – 15

By splitting the middle term,

x^{2} + 5x – 3x – 15

= x(x + 5) – 3(x + 5)

= (x – 3) (x + 5)

⇒ x = 3 or x =−5

## Video Lesson

### Condition for Common Roots

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