# The Fundamental Theorem of Algebra

The fundamental theorem of algebra says that every polynomial has at least one complex number as a solution.

### Table of Contents

- Introduction
- Complex Numbers
- Addition of Complex Numbers
- Example 1
- Example 2
- Subtraction of Complex Numbers
- Example 3
- Example 4
- Multiplying Complex Numbers
- Example 5
- Example 6
- Division of Complex Numbers
- Example 7
- Example 8
- Quadratic Formula
- Example 9
- Example 10
- Fundamental Theorem of Algebra
- Summary
- Thanks For Reading

**Fundamental Theorem Of Algebra**

Until this point, we have tried several ways to solve equations. You may have noticed there are some types that we can’t solve by using the previous methods. These equations do not have real zeros, their graphs do not cross the x-axis at any point.

That is why they are difficult to deal with. We end up with negative values for solutions. For practical answers, what does a negative value mean? It does not mean a lot. Therefore, we have the concept of complex numbers.

Complex numbers were invented by mathematicians a long time ago in order to deal with these situations. They allow us to deal with a negative quantity in a way that makes sense. People are most familiar with this concept as the imaginary number, “i”.

**Complex Numbers**

When a graph does not have any x-intercepts, it does not have any actual solutions. When a graph does not have any proper solutions, we have to use the imaginary number to make sense of it. So, we can look at complex numbers in a couple of different ways.

\[i = \sqrt{-1}\]

\[i^2 = -1\]

This means we can rewrite solutions as complex numbers. Previously, we would not have a solution. However, with complex numbers now defined, we have solutions that make sense.

We usually write complex numbers in standard form, (a + bi). The (a) part is the real number, while the (bi) is the imaginary number.

**Addition of Complex Numbers**

The addition of complex numbers is easily done. You are combining the real and imaginary parts of the expression. The process is the same for subtraction.

**Example 1 **

**Add: \(3i + 5i\)**

Add the two parts together.

\[8i\]

**Example 2**

**Add: **\((3+i) + (-5-2i)\)

Add the real and imaginary parts separately.

\[(-2-i)\]

**Subtraction of Complex Numbers**

Again, you are just combining the real and imaginary parts. Subtract the real and imaginary sections one by one.

**Example 3**

**Subtract: **\((12-7i) - (-1+9i)\)

In this example, you want to subtract the real and imaginary parts.

\[(13-16i)\]

**Example 4**

**Subtract: **\((3) - (4-6i)\)

Subtract the real and imaginary parts.

\[(-1+6i)\]

**Multiplying Complex Numbers**

We multiply complex number expressions together when needed. Multiply each part of the expression into the other parts. Then you combine terms and put it into the standard form.

**Example 5 **

**Multiply: **\((2) * (2+4i)\)

Multiply all the terms together.

\[(4+8i)\]

**Example 6**

**Multiply: **\((1+i) * (2-3i)\)

Multiply all the terms together.

\[2+2i-3i-3i^2\]

\[2-i-3(-1)\]

\[(5-i)\]

**Division of Complex Numbers**

We accomplish division of complex numbers, much like division of other algebraic expressions. For imaginary expressions, you multiply by the conjugate of the denominator. Combine all the like terms in order to simplify as much as possible. Then put the final version into standard form to finish.

**Example 7**

**Divide: **\(\frac{1}{1+i}\)

Multiply the expression by the conjugate of itself.

\[\frac{1}{1+i} * \frac{1-i}{1-i}\]

\[\frac{1-i}{1-i+i-i^2}\]

\[\frac{1-i}{2}\]

\[\frac{1}{2} - \frac{1}{2}i\]

**Example 8**

**Divide: **\(\frac{4+i}{5-i}\)

Multiply the expression by its conjugate.

\[\frac{4+i}{5-i} * \frac{5+i}{5+i}\]

\[\frac{20+4i+5i+i^2}{25+5i-5i-i^2}\]

\[\frac{19+9i}{26}\]

**Quadratic Formula**

So far, we have only seen the most simple expressions, the (a + bi) form. However, when presented with a quadratic equation, we need to approach it in another way. This is where the quadratic formula comes in. This may not be a surprise, it is a very useful tool for us.

**Example 9 **

**Solve: **\(x^2+5=0\)

Use the quadratic formula.

\[\frac{-5 \pm \sqrt{25}}{2}\]

\[\frac{-5 \pm 5}{2}\]

\[x=0, -5\]

**Example 10 **

**Solve: **\(5x^2+1 = 3x^2\)

Put it in quadratic form.

\[-5x^2+3x-1=0\]

Again, use the quadratic formula.

\[\frac{-3 \pm \sqrt{9-(4)(-5)(-1)}}{(2)(-5)}\]

\[\frac{-3 \pm \sqrt{-11}}{-10}\]

\[\frac{-3 \pm i \sqrt{11}}{10}\]

\[\frac{-3}{10} \pm \frac{i \sqrt{11}}{10}\]

**Fundamental Theorem of Algebra**

The previous sections are leading us to this important concept. You have probably noticed the expressions and equations all have complex zeros. This is because, when looking at equations, almost everything can have complex and imaginary parts.

This is the fundamental theorem of algebra. It means that every polynomial, with a degree equal to or greater than one, will have at least one complex zero.

Carl Gauss proved this theorem in 1797 in his doctoral thesis. This theorem says that every polynomial can be completely factored if we can use complex numbers.

**Summary**

In this section, we defined the imaginary unit and the resulting set of complex numbers. We concluded that every real number is also a complex number. Complex numbers are useful in the standard form, so we can work with them easier. We then concluded that every polynomial has at least one complex zero when using imaginary units. With this realization, we know we can completely factor every polynomial when using complex numbers.

**Thanks For Reading**

Thank you for reading this, I really appreciate it.

If you would like to join my newsletter, you can do so here:

If you need a suggestion on what to read next, then try these Math articles: