# Fundamental Theorem of Algebra

The "Fundamental Theorem of Algebra" is **not** the start of algebra or anything, but it does say something interesting about polynomials:

Any polynomial of degree **n** ... has **n** roots

but you may need to use complex numbers

Let me explain:

A Polynomial looks like this:

example of a polynomial this one has 3 terms |

The Degree of a Polynomial with one variable is ...

... the largest exponent of that variable.

A "root" (or "zero") is where the **polynomial is equal to zero**.

So, a polynomial of degree 3 will have 3 roots (places where the polynomial is equal to zero). A polynomial of degree 4 will have 4 roots. And so on.

### Example: what are the roots of **x**^{2} - 9?

^{2}- 9

**x ^{2} - 9** has a degree of 2, so there will be 2 roots.

Let us solve it. We want it to be equal to zero:

**x**

^{2}- 9 = 0First move the -9 to the other side:

**x**

^{2}= +9Then take the square root of both sides:

**x = ±3**

So the roots are **-3** and **+3**

And there is something else of interest:

A polynomial **can be rewritten like this**:

_{1}) are called

**Linear Factors**, because they make a line when you plot them.

### Example: **x**^{2} - 9

^{2}- 9

The roots are **r _{1} = -3** and

**r**(as we discovered above) so the factors are:

_{2}= +3^{2}- 9 =

**(x+3)(x-3)**

(in this case **a** is equal to **1** so I didn't put it in)

The Linear Factors are** (x+3)** and **(x-3)**

So when you know the **roots** you can then write down the **factors**.

Here is another example:

### Example: 3x^{2} - 12

Let us find the roots: We want it to be equal to zero:

^{2}- 12 = 0

3 and 12 have a common factor of 3:

^{2}- 4) = 0

We can solve **x ^{2} - 4** by moving the

**-4**to the right and taking square roots:

^{2}= 4

And so the factors are:

3x^{2} - 12 = 3(x+2)(x-2)

Likewise, when we know the **factors** of a polynomial we also know the **roots**.

### Example: 3x^{2 }- 18x^{ }+ 24

It is degree 2 (the largest exponent is 2) so there will be 2 factors.

3x^{2 }- 18x^{ }+ 24 = a(x-r_{1})(x-r_{2})

I just happen to know this is the factoring:

3x^{2 }- 18x^{ }+ 24 = 3(x-2_{})(x-4_{})

And so the roots (zeros) are:

- +2
- +4

Let us check those roots:

3(2)^{2 }- 18(2)^{ }+ 24 = 12 - 36 + 24 = **0**

3(4)^{2 }- 18(4)^{ }+ 24 = 48 - 72 + 24 = **0**

Yes! The polynomial is zero at x = +2 and x = +4

## Complex Numbers

But sometimes you need to use Complex Number to make the polynomial equal to zero.

A Complex Number is a combination of a Real Number and an Imaginary Number

And here is an example:

### Example: x^{2}-x+1

Can we make it equal to zero?

x^{2}-x+1 = 0

Using the Quadratic Equation Solver the answer (to 3 decimal places) is:

0.5 - 0.866i |
and | 0.5 + 0.866i |

They are complex numbers! But they still work.

And so the factors are:

x^{2}-x+1 = ( x - (0.5-0.866**i **) )( x - (0.5+0.866**i **) )

## Complex Pairs

So the roots r_{1}, r_{2}, ... etc may be Real or Complex Numbers.

But there is something interesting...

Complex Roots **always come in pairs**!

You saw that in our example above:

### Example: x^{2}-x+1

Had these roots:

0.5 - 0.866i |
and | 0.5 + 0.866i |

The pair are actually complex conjugates (where you **change the sign in the middle**) like this:

Always in pairs? Yes (unless the polynomial has complex coefficients, but we are only looking at polynomials with real coefficients here!)

So you either get:

**no**complex roots**2**complex roots**4**complex roots,- etc

And **never** 1, 3, 5, etc.

Which means you automatically know this:

Degree | Roots | Possible Combinations |
---|---|---|

1 | 1 | 1 Real Root |

2 | 2 | 2 Real Roots, or 2 Complex Roots |

3 | 3 | 3 Real Roots, or 1 Real and 2 Complex Roots |

4 | 4 | 4 Real Roots, or 2 Real and 2 Complex Roots, or 4 Complex Roots |

etc | etc! |

And so:

When the degree is odd (1, 3, 5, etc) there will be **at least one real root** ... guaranteed!

### Example: 3x-6

The degree is 1.

**There will be one real root**

At +2 actually:

:

You can actually see that it **must go through the x-axis** at some point.

## But Real is also Complex!

I have been saying "Real" and "Complex", but Complex Numbers do **include** the Real Numbers.

So when I say there are *"2 Real, and 2 Complex Roots"*, I should be saying something like *"2 Purely Real (no Imaginary part), and 2 Complex (with a non-zero Imaginary Part) Roots"* ...

... but that is a lot of words that sound confusing ...

... so I hope you don't mind my (perhaps too) simple language.

## Don't Want Complex Numbers?

If you don't want Complex Numbers, just multiply any pair of complex roots together:

(a + b**i**)(a - b**i**) = a^{2} + b^{2}

You get a Quadratic Equation with no Complex Numbers ... it is purely Real.

That type of Quadratic (where you can't "reduce" it any further without using Complex Numbers) is called an **Irreducible Quadratic**.

And remember that simple factors like (x-r_{1}) are called **Linear Factors**

So a polynomial can be factored into all Real values using:

**Linear Factors**, and**Irreducible Quadratics**

### Example: x^{3}-1

x^{3}-1 = (x-1)(x^{2}+x+1)

It has been factored into:

- 1 linear factor: (x-1)
- 1 irreducible quadratic factor: (x
^{2}+x+1)

To factor (x^{2}+x+1) further we need to use Complex Numbers, so it is an "Irreducible Quadratic"

## How do we know if the Quadratic is Irreducible?

Just calculate the "discriminant": b^{2} - 4ac

(Read Quadratic Equations to learn more about the discriminant.)

When **b ^{2} - 4ac** is negative, the Quadratic has Complex solutions,

and so is "Irreducible"

### Example: 2x^{2}+3x+5

a = 2, b = 3, and c = 5:

**b ^{2} - 4ac** = 3

^{2}- 4×2×5 = 9-40 =

**-31**

The discriminant is negative, so it is an "Irreducible Quadratic"

## Multiplicity

Sometimes a factor appears more than once. That is its **Multiplicity**.

### Example: x^{2}-6x+9

x^{2}-6x+9 = (x-3)(x-3)

"(x-3)" appears twice, so the root "3" has **Multiplicity of 2**

You need to include the **Multiplicities** when you say a polynomial of degree **n** has **n** roots.

### Example: x^{4}+x^{3}

There **should be** 4 roots (and 4 factors), right?

Factoring is easy, just factor out x^{3}:

x^{4}+x^{3} = x^{3}(x+1) = x·x·x·(x+1)

there are 4 factors, with "x" appearing 3 times.

But there seem to be only 2 roots, at **x=-1** and **x=0**:

But counting Multiplicities there are actually 4:

- "x" appears three times, so the root "0" has a
**Multiplicity of 3** - "x+1" appears once, so the root "-1" has a
**Multiplicity of 1**

Total = 3+1 = 4

## Summary

- A polynomial of degree
**n**has**n**roots (where the polynomial is zero) - A polynomial can be factored like:
**a(x-r**_{1})(x-r_{2})where r**...**_{1}, etc are the roots - Roots may be
**Complex Numbers** - Complex Roots
**always come in pairs** - Multiplying a Complex pair gives an
**Irreducible Quadratic** - So a polynomial can be factored into all real factors which are either:
**Linear Factors**or**Irreducible Quadratics**

- Sometimes a factor appears more than once. That is its
**Multiplicity**.