To show that the Fundamental Theorem of Algebra is true for the quadratic polynomial \( x^2 - 16x + 39 = 0 \) through solving by factoring, we can start by factoring the quadratic.
We need to find two numbers that multiply to \( 39 \) (the constant term) and add up to \( -16 \) (the coefficient of \( x \)).
The numbers that meet these criteria are \( -3 \) and \( -13 \), since: \[ (-3) + (-13) = -16 \quad \text{and} \quad (-3) \times (-13) = 39. \]
Using these numbers, we can factor the quadratic: \[ x^2 - 16x + 39 = (x - 3)(x - 13) = 0. \]
Now, we can solve for \( x \) by setting each factor equal to zero:
- \( x - 3 = 0 \) gives \( x = 3 \).
- \( x - 13 = 0 \) gives \( x = 13 \).
Thus, the solutions to the equation \( x^2 - 16x + 39 = 0 \) are \( x = 3 \) and \( x = 13 \).
Now, let's determine the nature of the solution set. Both solutions \( 3 \) and \( 13 \) are rational numbers. Therefore, the correct description of the solution set from the given options is:
There are two rational solutions.