## Polynomials

#### Knowing Polynomials

The word Polynomial comes from the Greek words where Poly means ‘Many’ and Nominal means ‘Terms’. So altogether it means “many terms”. Polynomial is a mathematical expression that consists of variables, constants, and exponents.

It is an algebraic expression in which the power of the variable is a non-negative integer(whole numbers).

**Note**:-

- Power can not be negative or in fraction form.
- The variable should not be present in the denominator.

Ex: $\sqrt x= x^{1/2}$

$1/{(𝒙^4 + 1)}$, here variable is in denominator

The above examples are not a polynomial.

Some examples of Polynomials are :

- $𝒙^3 + 2𝒙 + 4$
- $\sqrt 2𝒙^2$
- $𝒙^5/5$

#### Standard Form

$P(x) = a_n x^n + a_{n-1} x^{n -1} +a_{n-2} x^{n-2} + ………………. + a_1 x + a_0 $

Where $a_n , a_{n-1} , a{n-2} , ……………………, a_1, a_0$ are called coefficients of $x^n, x^{n-1}, x^{n-2}, ….., x$ and constant term respectively and it should belong to real number (⋲ R).

#### Real-life application

Engineers usually design the curves of the roller coaster by making use of the concepts in polynomials. Similarly, biologists study the expansion of bacterial colonies by making use of certain concepts of polynomials.

#### Zero polynomial

The constant polynomial 0 is called Zero Polynomial. In other words, it is a polynomial where all variable’s coefficients are equal to zero. The degree of zero polynomial is not defined.

Classifications of Polynomials

Polynomials are classified into two types based on

- Number of terms in a polynomial
- Degree of the polynomial

#### Classifications on number of terms

The terms of polynomials are the parts of the expression that are generally separated by “+” or “-” signs, based on the sign of addition and subtraction we will classify into different terms.

- Monomial: having only one term

**Eg**– $ 𝒙, 2𝒙^2, – 5𝒙^4 $

- Binomial: polynomials having only two(unlike)terms.

**Eg** – $ 𝒙+1, 𝒙^3+3𝒙, y^45-y^3 $

- Trinomial: polynomials having only three(unlike) terms.

**Eg**– $(𝒙^2+2𝒙+6), (x^3 +7𝒙^2 +𝛑)$

**Note **

- If there are 4 unlike terms or 5 unlike terms… They are all known as polynomials only.
- Monomials, binomials, and trinomials are polynomials only.

#### Classification on the degree of Polynomials

The degree of the polynomial is the highest power of the variable. The polynomial equation having one variable which has the largest exponent is called the degree of the polynomial.

**Eg**:- 2𝒙, Here the degree is 1.

$𝒙^2+3𝒙+5$, Here the degree is 2

$5 = 5 x 1$

$ = 5 x (𝒙^0)$

We know that $𝒙^0$ is 1, here the degree is 0.

Classification is based on the degree of a polynomial.

Linear Polynomial: a polynomial of degree one

**Eg**: $5𝒙, 10𝒙 +8$

Quadratic Polynomial: a polynomial of degree two

**Eg**: $ 𝒙^2-3𝒙+2, 5𝒙^2, 8𝒙^2+7𝒙+𝝅$

Cubic Polynomial: a polynomial of degree three

**Eg**: $𝒙^3-5𝒙^2+78, 45𝒙^3$

Constant Polynomial: a polynomial of degree zero

**Eg**: 8

Since $ 8 * 𝒙^0$, here the degree of a polynomial is zero.

#### Polynomials in One variable

- Concept and examples

Polynomials in one variable are those expressions in which there is only one variable present. To determine whether a polynomial is in one variable, we just have to see how many variables are present in the expression. If there is only one variable, let’s say x in the polynomial, we consider it as a polynomial in one variable.

An example of a polynomial with one variable is

$x^2+3x-12, $

$5x-2$

$x^3$

**Example**: Solve $3x – 15$

Solution:

First, make the equation 0. So,

$3x – 15 = 0$

$⇒ 3x = 15$

$⇒ x = 15/3$

Or, $x = 5.$

Thus, the solution of $3x-15$ is $x = 5.$

**Example**: Solve $2x – 8$

Solution:

First, make the equation 0. So,

$2x – 8 = 0$

$⇒ 2x = 8$

$⇒ x = 8/2$

Or, $x = 4.$

Thus, the solution of $2x-8$ is $x = 4$.

### Zeros of a polynomial

**Terminology and Understanding**

Zeros of a polynomial can be defined as the points where the polynomial becomes zero as a whole. A polynomial having a value of zero (0) is called a zero polynomial.

For finding zeros of a polynomial:

- Equate the polynomial to zero.
- Solve and find the value of the variable.

**Eg**: Find zeros of the following polynomials:

- $x + 5$

Solution: $x + 5 = 0$

$ x= 0 – 5$

$ x= -5$

2. Verify whether 2 and 0 are zeroes

of polynomial $x2-2x$

Solution: $P(x) = x^2-2x$

At $x = 2 : P(2) = 2^2 – 2*2$

$ = 4 – 4 = 0$

∴ 2 is a zero of a polynomial.

At $x = 0 : P(0) = 0^2 – 2* 0$

$= 0 – 0 = 0$

∴ 0 is a zero of a polynomial.

3) Check whether 2 and -2 are zeroes

of the polynomial $x + 2.$

Solution: $p(x) = x + 2$

At $x = 2 : p(2) = 2 + 2$

$= 4 +/0$

∴ 2 is not zero of a polynomial.

At $x = -2 : p(-2) = -2 + 2$

$ = 0$

∴ -2 is a zero of a polynomial.

### Remainder Theorem

#### Remainder Theorem with proo**f**

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).

It is applicable if the divisor is a linear polynomial.

ie, $ d(x) = x – a $

* We know that degree of remainder < degree of divisor*

Since here divisor is a linear polynomial, degree of remainder < 1.

∴ Degree of remainder = 0.

∴ Remainder (r) will be a constant number.

#### Applications of the Remainder theorem

The Polynomial Remainder Theorem allows us to determine whether a linear expression is a factor of a polynomial expression or not easily.

### Factor Theorem

If p(x) is a polynomial of degree [n ≥ 1] and a is any real number, then

(i) x-a is a factor of p(x), if p(a)=0, and

(ii) p(a)=0, if x-a is a factor of p(x).

It is an application of the remainder theorem.

When a number is divided by its factor; then the remainder will be 0. Similar is the case with polynomials as well.

*Factorisation of Quadratic Polynomials by splitting the middle term method*

Step 1: Write the given quadratic polynomial in standard form $a𝑥^2 + bx + c $

Step 2: Find the product of a and c.

Step 3: List down all factors of ac in pairs.

Step 4: Select a pair of factors such that their sum is ‘b’.

**Eg**: Factorise $2𝑥^2-8x+6$ by splitting the middle term.

Solution: Given the polynomial $2𝑥^2-8x+6$. Now to make the calculation

simple take the common term outside hence we get;

$2(𝑥^2 -4x + 3)$

So now consider the polynomial $x^2-4x+3$, we get $a = 1; b = -4; c = 3$

By using the steps mentioned above:

$ac= 1 x 3 = 3$

$b= -1-4 = -4$

i.e we have to choose the numbers such that on multiplying we should get 3 and on adding we should get -4. So here if we choose the -1 and -3, on adding we get -4, and on multiplying we get 3.

$x2-4x+3$

$= x^2-x-3x+3$

$= (x2-x)-(3x-3)$

$= x(x-1)-3(x-1)$

$= (x-1)(x-3)$

Hence the factors obtained will be

$2x^2-8x+6 $

$= 2(x-1)(x-3)$