To show that the Fundamental Theorem of Algebra is true for the quadratic polynomial \(x^2 - 16x + 39 = 0\) through solving by factoring, we first need to factor the quadratic expression.
We want to factor \(x^2 - 16x + 39\) into the form \((x - p)(x - q) = 0\), where \(p\) and \(q\) are the roots of the equation.
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To factor the quadratic, we can look for two numbers that multiply to \(39\) (the constant term) and add to \(-16\) (the coefficient of \(x\)). The factors of \(39\) are \(1, 3, 13, 39\).
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Upon inspection, we notice that the numbers \(-3\) and \(-13\) fit this criteria because:
- \(-3 \times -13 = 39\)
- \(-3 + -13 = -16\)
Thus, we can write the equation as: \[ (x - 3)(x - 13) = 0 \]
- Setting each factor to zero gives us the solutions: \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x - 13 = 0 \quad \Rightarrow \quad x = 13 \]
Now we have found the solutions \(x = 3\) and \(x = 13\).
Analyzing the solution set:
- Both \(3\) and \(13\) are rational numbers.
- There are no non-real solutions since both solutions are real numbers.
Based on this, the correct statement that accurately describes the solution set is: There are two rational solutions.
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