Complex Conjugates 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.

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
is the answer.