# Complex Conjugates Theorem

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

## 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

The highest degree term is x^{3}.

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 x^{3}.

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) = 3^{2} + 1^{2}

Divide Complex Numbers

3^{2} + 1^{2} = 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])

= (x^{2} - 6x + 10).

So

(x - 2)(x - [3 + i])(x - [3 - i])

= (x - 2)(x^{2} - 6x + 10).

-2⋅x^{2} = -2x^{2}

-2⋅(-6x) = +12x

-2⋅(+10) = -20

So

(x - 2)(x^{2} - 6x + 10)

= x^{3} - 6x^{2} + 10x - 2x^{2} + 12x - 20.

-6x^{2} - 2x^{2} = -8x^{2}

+10x + 12x = +22x

So

x^{3} - 8x^{2} + 22x - 20

is the answer.