**2.5 The Fundamental Theorem of Algebra – Proved by Carl Friedrich Gauss**

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**
**

**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}**