Given that one solution to the equation \( x^3 + 2x^2 - 5x - 6 = 0 \) is \( x = -1 \), we can factor the polynomial using this root.
First, we can perform polynomial long division or synthetic division to divide \( x^3 + 2x^2 - 5x - 6 \) by \( x + 1 \).
Using synthetic division:
\[ \begin{array}{r|rrrr} -1 & 1 & 2 & -5 & -6 \ & & -1 & -1 & 6 \ \hline & 1 & 1 & -6 & 0 \ \end{array} \]
The coefficients from the division show that:
\[ x^3 + 2x^2 - 5x - 6 = (x + 1)(x^2 + x - 6) \]
Now, we can factor the quadratic \( x^2 + x - 6 \).
To find the roots of \( x^2 + x - 6 = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = 1 \), and \( c = -6 \).
Calculating the discriminant:
\[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (-6) = 1 + 24 = 25 \]
Since the discriminant is positive, the quadratic has two distinct real solutions:
\[ x = \frac{-1 \pm \sqrt{25}}{2 \cdot 1} = \frac{-1 \pm 5}{2} \]
Calculating the two solutions:
- \( x = \frac{-1 + 5}{2} = \frac{4}{2} = 2 \)
- \( x = \frac{-1 - 5}{2} = \frac{-6}{2} = -3 \)
Thus, the full solution set of the equation \( x^3 + 2x^2 - 5x - 6 = 0 \) is:
\[ x = -1, 2, -3 \]
Now, let's summarize the solution set:
- There are three real solutions: \( -1, 2, -3 \).
Given the options:
- Option #1: It has one real solution and two non-real solutions. (Incorrect)
- Option #2: It has one non-real solution and two real solutions. (Incorrect)
- Option #3: It has three real solutions. (Correct)
Therefore, the correct answer is:
Option #3: It has three real solutions.