Question

Show that the Fundamental Theorem of Algebra is true for the quadratic polynomial x2−16x+39=0 through solving by factoring. Which of the following statements accurately describes the solution set?(1 point)%0D%0AResponses%0D%0A%0D%0AThere are two identical solutions.%0D%0AThere are two identical solutions.%0D%0A%0D%0AThere are two irrational solutions.%0D%0AThere are two irrational solutions.%0D%0A%0D%0AThere are two rational solutions.%0D%0AThere are two rational solutions.%0D%0A%0D%0AThere are two non-real solutions.

Answer 1

To solve the quadratic polynomial \(x^2 - 16x + 39 = 0\) by factoring, we first identify the coefficients:
- \(a = 1\) (coefficient of \(x^2\))
- \(b = -16\) (coefficient of \(x\))
- \(c = 39\) (constant term)

Next, we want to see if we can factor the quadratic into the form \((x - p)(x - q) = 0\), where \(p\) and \(q\) are the roots we need to find.

1. We find \(p\) and \(q\) such that:
\[
p + q = 16
\]
\[
pq = 39
\]

2. We can list pairs of factors of 39:
\[
(1, 39), (3, 13)
\]

The pair that adds up to 16 is \(3\) and \(13\). Thus we can set:
\[
p = 3, \quad q = 13
\]

3. Therefore, we can write the factorization as:
\[
(x - 3)(x - 13) = 0
\]

4. Setting each factor to zero gives us the solutions:
\[
x - 3 = 0 \Rightarrow x = 3
\]
\[
x - 13 = 0 \Rightarrow x = 13
\]

Thus, we have two distinct solutions: \(x = 3\) and \(x = 13\).

### Conclusion
The solutions are both rational numbers.

Since we are asked to determine which statement accurately describes the solution set, the correct response is:

**There are two rational solutions.**