# Polynomial Expressions

Polynomial expressions are algebraic expressions that have non-negative integers as their index. They are categorized based on their degree, which is the highest degree of the variables. Solving polynomial equations involves finding values of x that satisfy the equation. The process involves factoring the equation and setting each factor equal to zero to find the possible values of x. Some polynomial equations may have irrational solutions, which cannot be expressed as a ratio of integers.

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### What are polynomial expressions?

Algebraic expressions that have a non-negative integer as their index are known as polynomials. A non-negative integer means we can’t include rational numbers like β as well. Polynomials are categorized on the basis of their degree. The degree is nothing but the highest degree of the variables in the polynomial. An equation of the form π₯^n=p where p and n are integers denotes all simple higher-degree polynomials. Solving these types of equations is easy as it was with linear equations. Let’s take some of them one by one. For example,

π₯Β² = 4

The techniques that we learned to solve the equations were all related to the laws of equality. But both multiplicative and additive laws of equality cant be directly used in these types of equations. Solving this equation means finding out a square of what number is 4. A Square is the product of a number multiplied by itself. We know that 2 multiplied by itself is 4. Also, -2 multiplied by itself is also 4.

An equation of π₯^n=p form is said to have n possible values of π₯ that satisfy the equation.

How can we solve the equation π₯Β² = 4?

or, π₯Β² - 4 = 0

or, π₯Β² - 2Β² = 0

or, (π₯ - 2) (π₯ + 2) = 0 When we know that two numbers are multiplied to get 0, at least one of them must be 0. The product of two non-zero numbers cant is 0. Also, one fact is that whatever the value of π₯ be, both the terms (π₯ - 2) and (π₯ + 2) cant be 0 at the same time. If one of them is 0, the other isn’t. If (π₯ - 2) = 0 Then π₯ = 2

If (π₯ - 2) β 0, then (π₯ + 2) must be 0 (π₯ + 2) = 0 or, π₯ = -2

In another example, π₯Β³=27 What number when cubed becomes 27? π₯Β³-27=0

or, π₯Β³ - 3Β³ = 0

or, (π₯ - 3) (π₯Β² + 3π₯ +3Β²) = 0

or, (π₯ - 3) (π₯Β² + 3π₯ + 9) = 0

The same case here. At least one of the two expressions must be 0. Either π₯ - 3 = 0 or, π₯ = 3

If π₯ - 3 β 0, π₯Β² + 3π₯ + 9=0 The equation is a second-degree equation and should have given two solutions but this equation doesn’t give real numbers as solutions. We will learn about them later. Thus the only real solution we can get from the previous equation is π₯ = 3.

Letβs go back to another example of a second-degree equation. For example π₯Β²=8, No integer when squared gives 8. The square of 2 is 4 and 3 is 9 and 8 lies somewhere between the two. What we can be sure of is that the square root of 8 is somewhere between 2 and 3. Unlike the rational numbers between two integers, it cant be expressed in the form of a ratio or fraction of integers thus the solution is called as irrational.

β2, β3, β5, β4, etc. are some examples of irrational numbers and they mostly arise when a perfect root of a number cant be found.

Rational numbers are those which can be represented in the form of ratio p/q where both p and q are integers and q is not equal to 0. Irrational numbers are those which cant be expressed in the form of p/q.