**Polynomials** are as expressions which are composed of two algebraic terms. In this lesson, all the concepts of polynomials like its definition, terms and degree, types, functions, formulas and solution are covered along with solved example problems. Below is the list of topics covered in this lesson.

**Table of Contents:**

- Definition
- Degree
- Terms
- Types
- Properties and Theorems
- Equations
- Function
- Solving Polynomials
- Operations
- Examples

## What is a Polynomial?

A polynomial is defined as an expression which contains two or more algebraic terms. It is made up of two terms namely Poly (meaning “many”) and Nominal (meaning “terms.”). Polynomials are composed of:

- Constants such as 1, 2, 3, etc.
- Variables such as g, h, x, y, etc.
- Exponents such as 5 in x
^{5}etc.

## Degree of a Polynomial

The degree of a polynomials is defined as the highest degree of a monomial within a polynomial. Thus, a polynomial equation having one variable which has the largest exponent is called a degree of the polynomial.

Polynomial | Degree | Example |
---|---|---|

Constant or Zero Polynomial | 0 | 6 |

Linear Polynomial | 1 | 3x+1 |

Quadratic Polynomial | 2 | 4x^{2}+1x+1 |

Cubic Polynomial | 3 | 6x^{3}+4x^{3}+3x+1 |

Quartic Polynomial | 4 | 6x^{4}+3x^{3}+3x^{2}+2x+1 |

**Example: **Find the degree of the polynomial 6s^{4}+ 3x^{2}+ 5x +19

**Solution:**

The degree of the polynomial is 4.

## Terms of a Polynomial

The terms of polynomials are the parts of the equation which are generally separated by “+” or “-” signs. So, each part of a polynomial in an equation is a term. For example, in a polynomial, say, 2x^{2} + 5 +4, the number of terms will be 3. The classification of a polynomial is done based on the number of terms in it.

## Types of Polynomials

Polynomials are of 3 different types and are classified based on the number of terms in it. The three types of polynomials are:

**Monomial****Binomial****Trinomial**

These polynomials can be combined using addition, subtraction, multiplication, and division but is never division by a variable. A few examples of **Non Polynomials** are: 1/x+2, x^{-3}

### Monomial

A monomial is an expression which contains only one term. For an expression to be a monomial, the single term should be a non-zero term. A few examples of monomials are:

- 5x
- 3
- 6a
^{4} - -3xy

### Binomial

A binomial is a polynomial expression which contains exactly two terms. A binomial can be considered as a sum or difference between two or more monomials. A few examples of binomials are:

- – 5x+3,
- 6a
^{4}+ 17x - xy
^{2}+xy

### Trinomial

A trinomial is an expression which is composed of exactly three terms. A few examples of trinomial expressions are:

- – 8a
^{4}+2x+7 - 4x
^{2}+ 9x + 7

## Polynomial Properties and Theorems

Some of the important properties of polynomials along with some important polynomial theorems are as follows:

**Property 1: Division Algorithm**

If a polynomial P(x) is divided by a polynomial G(x) results in quotient Q(x) with remainder R(x), then,

**P(x) = G(x) ****• ****Q(x) + R(x)**

**Property 2: Bezout’s Theorem**

Polynomial P(x) is divisible by binomial (x – a) if and only if **P(a) = 0**.

**Property 3: Remainder Theorem**

If P(x) is divided by (x – a) with remainder r, then **P(a) = r**.

**Property 4: Factor Theorem**

A polynomial P(x) divided by Q(x) results in R(x) with zero remainders if and only if Q(x) is a factor of P(x).

**Learn More:** Factor Theorem

**Property 5: Intermediate Value Theorem**

If P(x) is a polynomial, and P(x) ≠ P(y) for (x < y), then P(x) takes every value from P(x) to P(y) in the closed interval [x, y].

**Learn More:** Intermediate Value Theorem

**Property 6:**

The addition, subtraction and multiplication of polynomials P and Q result in a polynomial where,

**Degree(P ± Q) ≤ Degree(P or Q)**

**Degree(P × Q) = Degree(P) + Degree(Q)**

**Property 7:**

If a polynomial P is divisible by a polynomial Q, then every zero of Q is also a zero of P.

**Property 8:**

If a polynomial P is divisible by two coprime polynomials Q and R, then it is divisible by (Q • R).

**Property 9:**

If P(x) = a_{0} + a_{1}x + a_{2}x^{2} + …… + a_{n}x^{n} is a polynomial such that deg(P) = n ≥ 0 then, P has at most “n” distinct roots.

**Property 10: Descartes’ Rule of Sign**

The number of positive real zeroes in a polynomial function P(x) is the same or less than by an even number as the number of changes in the sign of the coefficients. So, if there are “K” sign changes, the number of roots will be “k” or “(k – a)”, where “a” is some even number.

**Property 11: Fundamental Theorem of Algebra**

Every non-constant single-variable polynomial with complex coefficients has at least one complex root.

**Property 12:**

If P(x) is a polynomial with real coefficients and has one complex zero (x = a – bi), then x = a + bi will also be a zero of P(x). Also, x^{2} – 2ax + a^{2} + b^{2} will be a factor of P(x).

## Polynomial Equations

The polynomial equations are those expressions which are made up of multiple constants and variables. The standard form of writing a polynomial equation is to put the highest degree first then, at the last, the constant term. An example of a polynomial equation is:

b = a^{4} +3a^{3} -2a^{2} +a +1

## Polynomial Functions

A polynomial function is an expression constructed with one or more terms of variables with constant exponents. If there are real numbers denoted by a, then function with one variable and of degree n can be written as:

f(x) = a_{0}x^{n} + a_{1}x^{n-1 }+ a_{2}x^{n-2 }+ ….. + a_{n-2}x^{2 }+ a_{n-1}x + a_{n} |

## Solving Polynomials

Any polynomial can be easily solved using basic algebra and factorization concepts. To solve a polynomial equation, the first step is to set the right-hand side as 0. The explanation of a polynomial solution are explained for two types:

- Solving Linear Polynomials
- Solving Quadratic Polynomials

**Solving Linear Polynomials**

Getting the solution of linear polynomials is easy and simple. First, isolate the variable term and make the equation as equal to zero. Then solve as basic algebra operation. An example of finding the solution of a linear equation is given below:

**Example: **Solve 3x – 9

**Solution:**

First, make the equation as 0. So,

3x – 9 = 0

⇒ 3x = 9

⇒ x = 9/3

Or, x = 3.

Thus, the solution of 3x-9 is x = 3.

**Solving Quadratic Polynomials**

To solve a quadratic polynomial, first, rewrite the expression in the ascending order of degree. Then, equate the equation and perform polynomial factorization to get the solution of the equation. An example to find the solution of a quadratic polynomial is given below for better understanding.

**Example: **Solve 3x^{2} – 6x + x^{3} – 18

**Solution:**

First, arrange the polynomial in the ascending order of degree and equate to zero.

⇒ x^{3 }+ 3x^{2} -6x – 18 = 0

Now, take the common terms.

x^{2}(x+3) – 6(x+3)

⇒ (x^{2}-6)(x+3)

So, the solutions will be x =-3 and

x^{2} = 6

Or, x = √6

**More Polynomials Related Resources:**

## Polynomial Operations

There are four main polynomial operations which are:

- Addition of Polynomials
- Subtraction of Polynomials
- Multiplication of Polynomials
- Division of Polynomials

Each of the operations on polynomials are explained below using solved examples.

### Addition of Polynomials

To add polynomials, always add the like terms i.e. the terms having the same variable and power. The addition of polynomials always results in a polynomial of the same degree. For example,

**Example:** Find the sum of two polynomials: 5x^{3}+3x^{2}y+4xy−6y^{2}, 3x^{2}+7x^{2}y−2xy+4xy^{2}−5

**Solution:**

First, combine the like terms while leaving the unlike terms as they are. Hence,

(5x^{3}+3x^{2}y+4xy−6y^{2})+(3x^{2}+7x^{2}y−2xy+4xy^{2}−5)

= 5x^{3}+3x^{2}+(3+7)x^{2}y+(4−2)xy+4xy^{2}−6y^{2}−5

= 5x^{3}+3x^{2}+10x^{2}y+2xy+4xy^{2}−6y^{2}−5

### Subtraction of Polynomials

Subtracting polynomials is similar to addition, the only difference being the type of operation. So, subtract the like terms to obtain the solution. It should be noted that subtraction of polynomials also result in a polynomial of the same degree.

**Example:** Find the difference of two polynomials: 5x^{3}+3x^{2}y+4xy−6y^{2}, 3x^{2}+7x^{2}y−2xy+4xy^{2}−5

**Solution:**

First, combine the like terms while leaving the unlike terms as they are. Hence,

(5x^{3}+3x^{2}y+4xy−6y^{2})-(3x^{2}+7x^{2}y−2xy+4xy^{2}−5)

= 5x^{3}-3x^{2}+(3-7)x^{2}y+(4+2)xy-4xy^{2}−6y^{2}+5

= 5x^{3}-3x^{2}-4x^{2}y+6xy-4xy^{2}−6y^{2}+5

### Multiplication of Polynomials

Two or more polynomial when multiplied always result in a polynomial of higher degree (unless one of them is a constant polynomial). An example of multiplying polynomials is given below:

**Example:** Solve (6x−3y)×(2x+5y)

**Solution:**

⇒ 6x ×(2x+5y)–3y × (2x+5y) ———- Using distributive law of multiplication

⇒ (12x^{2}+30xy) – (6yx+15y^{2}) ———- Using distributive law of multiplication

⇒12x^{2}+30xy–6xy–15y^{2 }—————– as xy = yx

Thus, (6x−3y)×(2x+5y)=12x^{2}+24xy−15y^{2}

### Division of Polynomials

Division of two polynomial may or may not result in a polynomial. Let us study below the division of polynomials in details. To divide polynomials, follow the given steps:

**Polynomial Division Steps:**

If a polynomial has more than one term, we use long division method for the same. Following are the steps for it.

- Write the polynomial in descending order.
- Check the highest power and divide the terms by the same.
- Use the answer in step 2 as the division symbol.
- Now subtract it and carry down the next term.
- Repeat step 2 to 4 until you have no more terms to carry down.
- Note the final answer including remainder in the fraction form (last subtract term).

## Polynomial Examples

**Example:**

Given two polynomial 7s^{3}+2s^{2}+3s+9 and 5s^{2}+2s+1.

Solve these using mathematical operation.

**Solution:**

**Given polynomial:**

7s^{3}+2s^{2}+3s+9 and 5s^{2}+2s+1

**Polynomial Addition:** (7s^{3}+2s^{2}+3s+9) + (5s^{2}+2s+1)

= 7s^{3}+(2s^{2}+5s^{2})+(3s+2s)+(9+1)

= 7s^{3}+7s^{2}+5s+10

Hence, addition result in a polynomial.

**Polynomial Subtraction: **(7s^{3}+2s^{2}+3s+9) – (5s^{2}+2s+1)

= 7s^{3}+(2s^{2}-5s^{2})+(3s-2s)+(9-1)

= 7s^{3}-3s^{2}+s+8

Hence addition result in a polynomial.

**Polynomial Multiplication:**(7s^{3}+2s^{2}+3s+9) × (5s^{2}+2s+1)

= 7s^{3} (5s^{2}+2s+1)+2s^{2} (5s^{2}+2s+1)+3s (5s^{2}+2s+1)+9 (5s^{2}+2s+1))

= (35s^{5}+14s^{4}+7s^{3})+ (10s^{4}+4s^{3}+2s^{2})+ (15s^{3}+6s^{2}+3s)+(45s^{2}+18s+9)

= 35s^{5}+(14s^{4}+10s^{4})+(7s^{3}+4s^{3}+15s^{3})+ (2s^{2}+6s^{2}+45s^{2})+ (3s+18s)+9

= 35s^{5}+24s^{4}+26s^{3}+ 53s^{2}+ 21s +9

**Polynomial Division: (**7s^{3}+2s^{2}+3s+9) ÷ (5s^{2}+2s+1)

(7s^{3}+2s^{2}+3s+9)/(5s^{2}+2s+1)

This cannot be simplified. Therefore division of these polynomial do not result in a Polynomial.

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