Roots of Polynomials

A polynomial is a mathematical expression of the form:

$$ p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 $$

where \( a_n, a_{n-1}, \ldots, a_0 \) are constants that may belong to the integers \( \mathbb{Z} \), the rationals \( \mathbb{Q} \), the reals \( \mathbb{R} \), or the complex numbers \( \mathbb{C} \), and \( n \) is a non-negative integer.

A polynomial equation is an equation that takes the form \( p(x) = 0 \).

Here is an example of a polynomial equation:

$$ 3x^2 + 2x - 3 = 0 $$

A root of a polynomial equation \( p(x) = 0 \) is a number \( \alpha \) such that \( p(\alpha) = 0 \).

Some polynomial equations have solutions in the set of natural numbers \( \mathbb{N} \), others in the integers \( \mathbb{Z} \), the rationals \( \mathbb{Q} \), or the reals \( \mathbb{R} \).

However, every polynomial equation has at least one solution in the set of complex numbers \( \mathbb{C} \).

Polynomials over the Natural Numbers (\(\mathbb{N}\))

The natural numbers are \( \{1, 2, 3, \ldots\} \), commonly used for counting in everyday life.

For example, consider the equation:

$$ 4 + x = 9 $$

This equation has a solution in \( \mathbb{N} \): \( x = 5 \).

However, not all equations admit solutions in the natural numbers. For instance, the equation \( 9 + x = 4 \) has no solution in \( \mathbb{N} \), because 9 is already greater than 4, and adding any natural number to 9 only increases it further.

Polynomials over the Integers (\(\mathbb{Z}\))

To solve equations like \( 9 + x = 4 \), we extend our number system to include the integers \( \mathbb{Z} \), which also contain negative numbers: \( \{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\} \).

The equation \( 9 + x = 4 \) has a solution in \( \mathbb{Z} \): \( x = -5 \).

However, integers are still not sufficient for solving all polynomial equations. For instance, the equation \( 5x = 3 \) has no integer solution, as there is no integer \( n \) such that \( 5n = 3 \). To solve such equations, we must move to a broader number system.

Polynomials over the Rational Numbers (\(\mathbb{Q}\))

To address equations like \( 5x = 3 \), we turn to the rational numbers \( \mathbb{Q} \), which include all numbers that can be expressed as fractions: \( \left\{ \frac{m}{n} : m, n \in \mathbb{Z}, n \neq 0 \right\} \).

The equation \( 5x = 3 \) has a solution in \( \mathbb{Q} \): \( x = \frac{3}{5} \).

However, even the rationals are not sufficient for all polynomial equations. For example, the equation \( x^2 - 2 = 0 \) has no rational solution because \( \sqrt{2} \) is an irrational number.

Polynomials over the Real Numbers (\(\mathbb{R}\))

To solve equations like \( x^2 - 2 = 0 \), we work within the set of real numbers \( \mathbb{R} \), which includes all terminating and non-terminating decimal numbers.

The equation \( x^2 - 2 = 0 \) has two real solutions: \( x = \sqrt{2} \) and \( x = -\sqrt{2} \).

Real numbers provide solutions to many polynomial equations, but not all. For instance, the equation \( x^2 + 1 = 0 \) has no real solution, since no real number squared yields a negative value.

Polynomials over the Complex Numbers (\(\mathbb{C}\))

To solve equations such as \( x^2 + 1 = 0 \), we use complex numbers \( \mathbb{C} \), which include the imaginary unit \( i \), defined by the property \( i^2 = -1 \).

For example, \( x^2 + 1 = 0 \) has two complex roots: \( x = i \) and \( x = -i \).

The set of complex numbers contains all the previously discussed number systems - real, rational, integer, and natural - and it guarantees that every polynomial equation has at least one root.

This is the essence of one of the most important results in algebra, known as the Fundamental Theorem of Algebra, which states that every non-constant polynomial with complex coefficients has at least one complex root.