Question

Show that the Fundamental Theorem of Algebra is true for the quadratic polynomial x^2-16x+39=0 through solving by factoring. Which of the following statements accurately describes the solution set? (1 point) There are two rational solutions. There are two irrational solutions. There are two non-real solutions. There are two identical solutions.

Answer 1

To show that the Fundamental Theorem of Algebra is true for the quadratic polynomial \( x^2 - 16x + 39 = 0 \) through factoring, we will first attempt to factor the quadratic expression.

1. The given quadratic is \( x^2 - 16x + 39 \).
2. We are looking for two numbers that multiply to \( 39 \) (the constant term) and add up to \( -16 \) (the coefficient of the linear term).

Let's consider the factors of 39:
- The pairs of factors are \( (1, 39) \) and \( (3, 13) \).

Analyzing each pair:
- None of these pairs add up to \(-16\), but we can find the roots using the quadratic formula since direct factoring isn't viable.

To use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For our quadratic \( a = 1 \), \( b = -16 \), and \( c = 39 \), we substitute:

Calculating the discriminant:
\[
b^2 - 4ac = (-16)^2 - 4 \cdot 1 \cdot 39 = 256 - 156 = 100
\]

Now substituting back into the quadratic formula:
\[
x = \frac{16 \pm \sqrt{100}}{2 \cdot 1} = \frac{16 \pm 10}{2}
\]

This gives us two potential solutions:
1. \( x = \frac{16 + 10}{2} = \frac{26}{2} = 13 \)
2. \( x = \frac{16 - 10}{2} = \frac{6}{2} = 3 \)

Thus, the solutions of the equation \( x^2 - 16x + 39 = 0 \) are \( x = 13 \) and \( x = 3 \).

Now, we analyze the solution set. Both solutions are:
- Rational numbers (since both 13 and 3 can be expressed as fractions with an integer).

Therefore, the correct statement describing the solution set is:
**There are two rational solutions.**