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 x3.
So this function has at most 3 zeros.
So these are the zeros.
x = 2, 3 + i, 3 - i.
The highest degree term is x3.
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) = 32 + 12.
Complex conjugates
32 = 9
12 = 1
9 + 1 = 10
(3 + i) + (3 - i) = 6
(3 + i)(3 - i) = 10
So [x - (3 + i)][x - (3 - i)]
= x2 - 6x + 10.
Sum and product of the roots of a quadratic equation
So f(x) = (x - 2)(x2 - 6x + 10)
Solve (x - 2)(x2 - 6x + 10).
x(x2 - 6x + 10)
= x3 - 6x2 + 10x
-2(x2 - 6x + 10)
= -2x2 + 12x - 20
Multiplying polynomials
x3 = x3
-6x2 - 2x2 = -8x2
+10x + 12x = +22x
-20 = -20
So f(x) = x3 - 8x2 + 22x - 20.
