# Fundamental Theorem of Algebra

## The Essential Guide to the Fundamental Theorem of Algebra

Welcome to this lesson on the Fundamental Theorem of Algebra, a crucial concept in understanding how to factor and solve polynomials by identifying their roots. But before we dive into this fundamental theorem, let's quickly review some important definitions.

### Defining Important Terms

It's important to understand the following terms before we explore the theorem.

### Multiplicity

A polynomial p(x) has multiple roots at r, the multiplicity of r refers to how many times (x – r) occurs as a factor of p(x). We also call these repeated roots. For instance, p(x) = (x – r)3 means that the root r has a multiplicity of 3.

### Roots

The roots of a polynomial p(x) are values of the variable that make the equation p(x) = 0 true. We also refer to these as solutions, zeros, or x-intercepts.

### Standard Form and Degree of a Polynomial

A polynomial p(x) can be written as a sum of terms, where the coefficients are represented by and the variables with respective powers. The highest power of x in a polynomial with non-zero coefficients is known as the degree of the polynomial.

Now, let's move on to the theorem itself.

### The Essential Fundamental Theorem of Algebra

If a polynomial p(x) has a degree n ≥ 1, then it will have exactly n roots, including multiplicities and complex roots.

It's worth noting that n here refers to the highest degree of the polynomial. While we won't be proving this theorem in this lesson, it's crucial to understand how it can be applied when factoring and solving polynomials.

Additionally, the term "complex" in this context describes a root with a non-zero imaginary part, for example, a + bi, where a is real and b ≠ 0. As complex roots always appear in pairs, it means that a - bi is also a root of the polynomial.

### Using the Fundamental Theorem of Algebra

Now that we know what FTA is, let's see it in action. We'll work through a few problems to get a better understanding of how it can be used for factoring and solving polynomials. For simplicity, we'll refer to the Fundamental Theorem of Algebra as FTA.

### Identifying the Number of Roots in a Polynomial

With FTA, we can quickly determine the number of roots in a polynomial.

**Example 1:** Using FTA, find the number of roots for the polynomial **f(x) = x ^{4} + 2x^{3} + 3x^{2} + 4x + 5**

**Solution:** Since f(x) has a degree n = 4, according to FTA, it will have 4 solutions.

**Example 2:** Using FTA, find the number of roots for the polynomial **f(x) = x ^{7} - 3x^{5} + 4x^{3} + 2x^{2} - 1**

**Solution:** Since f(x) has a degree n = 7, according to FTA, it will have 7 solutions.

Therefore, from FTA, we can deduce that:

- A linear polynomial (degree 1) has one root
- A quadratic polynomial (degree 2) has two roots
- A cubic polynomial (degree 3) has three roots
- An nth degree polynomial has n roots

### Identifying the Zeros and Degree of a Polynomial

A polynomial p(x) can be written as a product of linear factors, where the roots of the polynomial are given by **r _{1}, r_{2}, ..., r_{n}**.

**Example 1:** Find the number of roots and their solutions for the factored polynomial **f(x) = (x + 2)(x - 3)(x - 5)**

**Solution:** By setting f(x) = 0 and using the Zero Product Property, we find that the roots are **-2, 3, 5**. Since there are 3 roots in total, FTA tells us that the polynomial must have a degree of 3.

**Example 2:** Find the number of roots and their solutions for the factored polynomial **f(x) = (x + 1)(x - 1)(x ^{2} + 4)**

**Solution:** Similarly, by setting f(x) = 0, we find that the roots of the polynomial are **-1, 1, 2i, -2i**. As there are 6 roots in total, FTA tells us that the polynomial must have a degree of 6.

### Factoring and Solving a Polynomial

We can use the Fundamental Theorem of Algebra to factor and solve polynomials, even if they can't be written as a product of linear factors.

**Example 1:** Factorize and solve the polynomial **f(x) = x ^{4} - 16**

**Solution:** First, we set f(x) = 0 and observe that this is a difference of two squares. From there, we can factor the polynomial as (x^{2} - 4)(x^{2} + 4), which can be further simplified to (x + 2)(x - 2)(x^{2} + 4). The solutions for this polynomial are x = ±2 and ±2i.

## Understanding Special Products and Factoring Polynomials

In algebra, we often encounter polynomials that can be rewritten using special products. For example, an expression in the form of **(x ^{2} - 4)(x^{2} + 4)** can be further simplified to

**(x - 2)(x + 2)(x**, allowing us to easily identify the solutions as

^{2}+ 4)**2, -2, 2i, -2i**. This type of polynomial has a degree of 4 and has 4 roots, including 2 real roots and 2 complex conjugate roots.

**Example 2:** Factoring and Solving a Polynomial

**Solution:** Let's take the polynomial **f(x) = x ^{3} - 4x^{2} + x - 4** as another example. By setting f(x) = 0, we can simplify it to

**(x**, giving us the zeros of

^{2}- 4x) + (x - 4) = x(x - 4) + 1(x - 4) = (x + 1)(x - 4)**-1, 4**. This polynomial has a degree of 3 and has 3 roots, including 1 real root and 2 complex conjugate roots.

While we have only looked at polynomials that can be factored into a product of linear factors so far, we may also encounter irreducible quadratics that cannot be simplified further. For instance, **x ^{2} + 4** and

**x**are examples of irreducible quadratics with two complex conjugate roots.

^{2}+ 9## Identifying Irreducible Quadratics Using the Discriminant

In cases where the irreducible quadratic cannot be factored into linear factors, we can use the discriminant of a quadratic polynomial to identify it. The discriminant provides a rule for determining the type of roots a quadratic has.

**The Discriminant of a Quadratic Equation**

The discriminant describes the roots of a quadratic polynomial and can be categorized into three cases:

**Case 1:**D > 0 - This means that the polynomial can be factored into linear factors, resulting in two distinct real roots.**Case 2:**D = 0 - In this case, the polynomial can be factored into multiplicity, resulting in one real repeated root.**Case 3:**D < 0 - This case results in an irreducible quadratic with two complex conjugate roots.

Let's apply this to an example and solve a polynomial using the quadratic formula.

## Solving an Irreducible Quadratic with the Quadratic Formula

**Solution**

Let's say we have the polynomial f(x) with the equation f(x) = x^{3} – 4x^{2} – 7x + 10 and need to factorize and solve it. By using the Fundamental Theorem of Algebra (FTA), we know that f(x) has a degree of 3, meaning there must be three solutions. After performing long division, we can factorize f(x) as (x + 2)(x – 5)(x – 1).

The discriminant in this case is **-7 ^{2} - 4(1)(10) = 9**, indicating that there are two complex conjugate roots. Therefore, we can use the quadratic formula to solve for the remaining two roots, resulting in one real root of x = 2, and two complex conjugate roots of x = -1 ± 2i.

## Building a Polynomial Using the FTA

In addition to factoring polynomials, we can also use the FTA to create a polynomial from a given statement. Let's look at two examples of how this can be done.

**Example 1:** Creating a Polynomial with Specific Zeros and Degree

**Solution**

To create a polynomial f(x) with zeros of 3 and -5, a degree of 3, and -5 having a multiplicity of 2, we can use the factors (x – 3) and (x + 5). Since the degree of f(x) is 3, we need one more factor, resulting in the completely factored form of f(x) as (x – 3)(x + 5)(x + a).

Since the multiplicity of -5 is 2, we need two factors of (x + 5), giving us a standard form of f(x) as (x – 3)(x + 5)(x + 5). By expanding using the FOIL method, we get the standard form of f(x) as x^{3} + 7x^{2} – 10x – 75.

**Example 2:** Creating a Polynomial with Complex Zeros

**Solution**

If the zeros of f(x) are -3, -4i, and 4i, we can use the factors (x + 3), (x^{2} + 16), and (x^{2} + 16). Expanding this using the FOIL method gives us the standard form of f(x) as x^{3} + 3x^{2} + 48x + 48.

## The Importance of Understanding the Fundamental Theorem of Algebra

- The Fundamental Theorem of Algebra states that a polynomial of degree n will have n roots when the polynomial is equal to 0.
- A polynomial of the form p(x) = an xn + ... + a1 x1 + a0 can be broken down into a product of linear factors in the form of p(x) = a(x – r1)(x – r2)...(x – rn).
- The roots of a polynomial can take on various forms, such as real numbers, multiplicities, or complex numbers. Complex number roots always appear in pairs of complex conjugates.
- A polynomial can be factored into a linear factor or an irreducible quadratic.
- A repeated root, also known as a multiplicity, is a factor that appears more than once in an expression.

## What is the Fundamental Theorem of Algebra?

The Fundamental Theorem of Algebra is a fundamental concept in mathematics that explains the relationship between a polynomial's degree and its roots.

## Why is the Fundamental Theorem of Algebra Important?

Understanding the Fundamental Theorem of Algebra has practical applications, such as being able to factorize and solve polynomials with ease.

## The Significance of the Fundamental Theorem of Linear Algebra

The Fundamental Theorem of Linear Algebra emphasizes that every polynomial has at least one root in the complex numbers, which can be represented as a linear factor.

## What are Roots in the Fundamental Theorem of Algebra?

The roots of a polynomial are values of the variable that make the equation true. These terms are also referred to as solutions, zeros, or x-intercepts.

## How Can You Apply the Fundamental Theorem of Algebra?

To utilize the Fundamental Theorem of Algebra, one must determine the polynomial's degree and coefficients and then utilize appropriate methods to factorize and solve it.