# Complex Conjugates Theorem

How to use the complex conjugates theorem to find the missing zeros: theorem, example, and its solution.

## Theorem

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

then its complex conjugate, (*a* - *bi*),

is also the zero of *f*(*x*).

Complex conjugates

So both [*x* - (*a* + *bi*)] and [*x* - (*a* - *bi*)]

are the factors of the *f*(*x*).

Factor theorem

## Example

The given zeros are 2 and 3 + *i*.

So, by the complex conjugate theorem,

the complex conjugate of (3 + *i*), (3 - *i*),

is also the zero of the function.

So *x* = 2, 3 + *i*, 3 - *i*.

The highest degree term is *x*^{3}.

So this function has at most 3 zeros.

So these are the zeros.

*x* = 2, 3 + *i*, 3 - *i*.

The highest degree term is *x*^{3}.

So, by the factor theorem,*f*(*x*) = (*x* - 2)[*x* - (3 + *i*)][*x* - (3 - *i*)].

Factor theorem

To expand [*x* - (3 + *i*)][*x* - (3 - *i*)],

find the sum and the product of the zeros.

Sum and product of the roots of a quadratic equation

Add the zeros.

Then (3 + *i*) + (3 - *i*) = 6.

Multiply the zeros.

Then (3 + *i*)(3 - *i*) = 3^{2} + 1^{2}.

Complex conjugates

3^{2} = 9

1^{2} = 1

9 + 1 = 10

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

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

So [*x* - (3 + *i*)][*x* - (3 - *i*)]

= *x*^{2} - 6*x* + 10.

Sum and product of the roots of a quadratic equation

So *f*(*x*) = (*x* - 2)(*x*^{2} - 6*x* + 10)

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

= *x*^{3} - 6*x*^{2} + 10*x*

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

= -2*x*^{2} + 12*x* - 20

Multiplying polynomials

*x*^{3} = *x*^{3}

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

+10*x* + 12*x* = +22*x*

-20 = -20

So *f*(*x*) = *x*^{3} - 8*x*^{2} + 22*x* - 20.