# Complex Conjugates Theorem

How to use the complex conjugates theorem to find the roots of an equation: theorem, 1 example, and its solution.

## Theorem

### Theorem

If [a + bi] is the zero of f(x),

then its conjugate [a - bi]
is also the zero of f(x).

This is the complex conjugates theorem.

## Example

### Solution

The highest degree term is x3.
Then there are 3 zeros.

The known zeros are
x = 2 and 3 + i.

Then, by the complex conjugates theorem,
the conjugate of 3 + i, 3 - i,
is also the zero.

So x = 2, 3 + i, 3 - i
are the zeros.

The zeros are
x = 2, 3 + i, 3 - i.

The highest degree term is x3.
The coefficient is 1.

So f(x) = (x - 2)(x - [3 + i])(x - [3 - i]).

Factor Theorem

Expand (x - [3 + i])(x - [3 - i]).

Instaed of directly expanding this,
use the sum and the product of the roots.

(3 + i) + (3 - i) = 6

The product of the roots is
(3 + i)(3 - i).

(3 + i)(3 - i) = 32 + 12

Formula: Divide Complex Numbers

32 + 12 = 9 + 1

9 + 1 = 10

Write the front factor (x - 2).

(3 + i) + (3 - i) = 6
(3 + i)(3 - i) = 10

Then
(x - [3 + i])(x - [3 - i])
= (x2 - 6x + 10).

So
(x - 2)(x - [3 + i])(x - [3 - i])
= (x - 2)(x2 - 6x + 10).

Expand (x - 2)(x2 - 6x + 10).

Multiply Polynomials

x⋅x2 = x3
x⋅(-6x) = -6x2
x⋅(+10) = +10x

-2⋅x2 = -2x2
-2⋅(-6x) = +12x
-2⋅(+10) = -20

So
(x - 2)(x2 - 6x + 10)
= x3 - 6x2 + 10x - 2x2 + 12x - 20.

-6x2 - 2x2 = -8x2
+10x + 12x = +22x

So
x3 - 8x2 + 22x - 20