**2.5 The Fundamental Theorem of Algebra –**

Proved by Carl Friedrich Gauss

Proved by Carl Friedrich Gauss

If

*f*(

*x*) is a polynomial of a degree “

*n*”, where n is greater than 0,

“

*f*” has at least one zero in the complex number system.

**I. Linear Factorization Theorem:**

If f(x) is a polynomial of degree “n” where “n is greater than zero”, “

*f*” has precisely “

*n*” linear factors.

f (x) = a

_{n}(x – c

_{1})( x – c

_{2})( x – c

_{3}) … ( x – c

_{n}) where

c

_{1}, c

_{2}, c

_{3}, … c

_{n}are complex numbers.

**Factors of a Polynomial:**

**Example 1:**f (x) = t

^{4}- 5t

^{3}+ 15t

^{2}- 45t + 54

4th degree

*n*= 4,

*n*is greater than zero is true

**Step 1:**find all the possible zeros of the function:

(factors of 54)/(factors of 1) = ±1, ± 2, ± 3, ±6, ± 9, ± 18, ± 27, ± 54

**Step 2:**Use synthetic division to find out which ones are factors

f (x) = (x – 2)(x – 3)(x

^{2}+ 9)

0 = (x – 2)(x – 3)(x

^{2}+ 9)

0 = x – 2

x = 2

0 = x – 3

x = 3

0 = x

^{2}+ 9

- 9 = x

^{2}

±√(-9) = x

± 3i = x

So to write this function in linear form:

f(x) = (x – 2)(x – 3)(x + 3i)(x – 3i)

So the following zeros of “

*f*” are:

*x*= {2, 3, 3i, and -3i}

**II. Complex Zeros Occur in Conjugate Pairs**

In the last example, “± 3

*i*”, the pair is 3

*i*and -3

*i*so we can conclude:

Let

*f*(

*x*) be a polynomial function that has real coefficients.

If

*a*+

*bi*, where

*b*≠ 0, is a zero of the function,

the conjugate

*a*–

*bi*is also a zero of the function.

Example: Your zeros are {2, 4 +

*i,*and 4 –

*i*}, find the polynomial function:

*f*(x) = (

*x*– 2)(

*x*– (4 +

*i*))(

*x*– (4 –

*i*))

= (

*x*– 2)((

*x*– 4) –

*i*)((x – 4) +

*i*)

= (

*x*– 2)( (

*x*– 4)

^{2}- (

*i*)

^{2})

= (

*x*– 2)(

*x*

^{2}- 8

*x*+ 16 – (-1))

= (

*x*– 2)(

*x*

^{2}- 8

*x*+ 17)

=

*x*

^{3}– 8

*x*

^{2}+ 17

*x*– 2

*x*

^{2}+ 16

*x*– 34

=

*x*

^{3}– 10

*x*

^{2}+ 33

*x*– 34

**III. If you are given one zero, can you find the rest?**

*g*(

*x*) = 4

*x*

^{3}+ 23

*x*

^{2}+ 34

*x*– 10 given zero:

**-3 +**

*i*Recall both conjugates:

– 3 +

*i*and – 3 –

*i*so

(

*x*+ 3 –

*i*) (

*x*+ 3 +

*i*)

= ((

*x*+ 3) –

*i*)((

*x*+ 3) +

*i*)

= ((

*x*+ 3)

^{2}- (

*i*)

^{2})

=

*x*

^{2}+ 6

*x*+ 9 – (-1)

=

*x*

^{2}+ 6

*x*+ 10

So using long division:

Every polynomial of degree

*n*is greater than zero with real coefficients can be written as the product of linear and quadratic factors with real coefficients, where the quadratic factors have no real zeros.

A quadratic factor with no real zeros is said to be irreducible over the real numbers.

x

^{2}+ 1 = (x +

*i*)(x –

*i*) is irreducible over the real numbers

x

^{2}- 2 = (x + √2)(x - √2) is irreducible over the rational numbers but reducible over the real numbers.

**Example:**The complex number 4

*i*is a zero of f(

*x*) =

*x*

^{4}+ 13

*x*

^{2}- 48.

Find the remaining zeros of

*f*(

*x*), and write it in its linear factorization.

Since 4

*i*is a zero, then -4

*i*is also a zero so

(

*x*-4

*i*)(

*x*+4

*i*) = x

^{2}- (4

*i*)

^{2}

=

*x*

^{2}- 16

*i*

^{2}

=

*x*

^{2}+ 16

therefore a factor of

*f*(

*x*) is

*x*

^{2}+ 16

*f*(

*x*) =

*x*

^{4}+ 13

*x*

^{2}- 48

= (

*x*

^{2}+ 16)(x

^{2}- 3)

= (

*x*+ 4

*i*)(

*x*– 4

*i*)(

*x*+ √3)(

*x*- √3)

**So the zeros are:**

*x*= {-4i, 4i, -√3, √3}

**Homework:**Pg. 170/ 57, 65, 69

*Pg. 187/ 1 – 5odd, 9 - 19odd, 33, 37, 41, 45, 53, 55, 61, 65