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Math 105: Precalculus Algebra14 chapters | 124 lessons | 12 flashcard sets

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Lesson Transcript

Instructor:
*David Liano*

In this lesson, you will learn what the Fundamental Theorem of Algebra says. You will also learn how to apply this theorem in determining solutions of polynomial functions.

This lesson will show you how to interpret the fundamental theorem of algebra. After completing this lesson, you will be able to state the theorem and explain what it means. Before we state the theorem, we will consider the following analogy.

Let's say your bank charges a fee every time you withdraw money from an automatic teller machine. If you withdraw money five times in a particular month, then you will expect five respective bank fees on that month's statement. Let's change this statement by using some mathematical lingo:

If you withdraw money *n* times in a particular month, then you will expect *n* respective bank fees on that month's statement.

5 withdrawals = 5 bank fees

The fundamental theorem of algebra is just as straightforward as this banking analogy.

The **fundamental theorem of algebra** states the following:

*A polynomial function f(x) of degree n (where n > 0) has n complex solutions for the equation f(x) = 0.*

Please note that the terms 'zeros' and 'roots' are synonymous with solutions as used in the context of this lesson.

That is pretty much it. Now, we should already know that polynomials can be described by their degree. For example, the polynomial *x*^3 + 3*x*^2 - 6*x* - 8 has a degree of 3 because its largest exponent is 3. The degree of a polynomial is important because it tells us the number of solutions of a polynomial.

The theorem does not tell us what the solutions are. It only tells us how many solutions exist for a given polynomial function.

So what good is that? First of all, it is important to understand underlying concepts of any math topics you are learning. In addition, the fundamental theorem of algebra has practical applications. For instance, if you need to find the solutions of a polynomial function, say, of degree 4, you know that you need to keep working until you find 4 solutions.

It is important to note that the theorem says **complex solutions**, so some solutions might be imaginary or have an imaginary part. Maybe we should do a quick review of complex numbers.

Complex numbers are in the form of *a* + *bi* (*a* and *b* are real numbers). The term *a* is the real part, and the term *bi* is the imaginary part. If *b* = 0, then the number is a real number.

Therefore, all real numbers are complex numbers. Let's look at a couple of examples:

In the complex number 2 + 3*i*, 2 is the real part and 3*i* is the imaginary part.

In the complex number 25 + 0*i*, 25 is the real part and 0*i* is the imaginary part. Because *b* = 0, the number simplifies to 25.

Before we look at some examples of polynomial functions, let's clarify the concept of **repeated solutions**. A polynomial function has repeated solutions if it has repeated factors.

A good way to show this is with the function *f(x)* = *x*^3. This function has a degree of 3, so based on our theorem, it has 3 solutions. We might see the three solutions better if we show the function in factored form: *f(x) = (x)(x)(x)*. Let's now make the function equal to zero: 0 = *(x)(x)(x)*. If any of the three factors equal zero, then the function equals zero. Therefore, the solutions are *x* = 0, *x* = 0, and *x* = 0. The solution of zero occurs 3 times.

Let's start with the polynomial function *f(x)* = *x*^2 + 9. In factored form, this function equals (*x* - 3*i*)(*x* + 3*i*).

This function has a degree of 2, so it has two solutions, which are *x* = 3*i* and *x* = -3*i*. These are imaginary solutions, so the graph of the function does not cross the *x*-axis. In other words, it has no *x*-intercepts. You might have noticed that the imaginary solutions are a **conjugate pair**. In fact, imaginary solutions to polynomial functions that have real numbers for coefficients always occur in conjugate pairs.

If *a* + *bi* (when *b* does not equal zero) is a solution of *f(x)* that is a polynomial with real coefficients, then its conjugate *a* - *bi* is also a solution of *f(x)*.

The graph of this function is shown below:

Our next function is *f(x)* = *x*^3 + 3*x*^2 - 6*x* - 8. Let's also look at the graph of the function.

It clearly crosses the *x*-axis three times, so all the solutions must be real solutions. In fact, the factored form of this function is *f(x)* = (*x* + 1)(*x* - 2)(*x* + 4). The solutions for this function are *x* = -1, *x* = 2, and *x* = -4. These solutions can also be determined by looking at where the graph crosses the *x*-axis. These must be the only solutions because the function has a degree of 3.

The final function that we will look at is *f(x)* = *x*^4 + 2*x*^3 - 2*x*^2 + 8. Let's look at the graph of this function.

Notice that this function touches the *x*-axis at *x* = -2. When a graph touches but does not cross the *x*-axis, it tells us that we have a repeated solution (in this case, *x* = -2 occurs twice). The graph does not cross the *x*-axis at any other points, so the other solutions must be imaginary.

Even though the same factor (*x* + 2) occurs twice, it still creates two solutions for the function. The other factors are clearly a conjugate pair of imaginary factors, as expected. Our four solutions are as follows:

*f(x)* = (*x* + 2)(*x* + 2)(*x* - (1 - *i*))(*x* - (1 + *i*))

This simplifies into: *x* = -2, *x* = -2, *x* = 1 - *i*, and *x* = 1 + *i*

The **fundamental theorem of algebra** simply states that the number of complex solutions to a polynomial function is equal to the degree of a polynomial function. Knowing this theorem gives you a good starting point when you are required to find the factors and solutions of a polynomial function. Also, do not forget about using graphs of polynomial functions to help you. They can show if the solutions are real and/or imaginary. Graphs can also provide evidence of repeated solutions.

As soon as you successfully work through this lesson, you could have the capability to:

- Comprehend the fundamental theorem of algebra
- Display your understanding of repeated solutions and complex solutions
- Apply the theorem when solving polynomial functions

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Math 105: Precalculus Algebra14 chapters | 124 lessons | 12 flashcard sets

- How to Add, Subtract and Multiply Polynomials 6:53
- Factoring Polynomials Using Quadratic Form: Steps, Rules & Examples 8:38
- How to Divide Polynomials with Long Division 8:05
- How to Use Synthetic Division to Divide Polynomials 6:51
- Remainder Theorem & Factor Theorem: Definition & Examples 7:00
- Dividing Polynomials with Long and Synthetic Division: Practice Problems 10:11
- Finding Rational Zeros Using the Rational Zeros Theorem & Synthetic Division 8:45
- Fundamental Theorem of Algebra: Explanation and Example 7:39
- Go to Polynomial Functions of a Higher Degree

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