To determine the number of solutions for the polynomial $x^3 - 6x^2 - 4x + 24 = 0$, we can analyze it based on the Fundamental Theorem of Algebra and factorization.

Step-by-Step Solution

1. Degree of the Polynomial: The polynomial $x^3 - 6x^2 - 4x + 24$ is a third-degree polynomial (the highest power of $x$ is 3). This means it can have up to 3 solutions (real or complex).

2. Finding Possible Real Roots: We can use the Rational Root Theorem to check for rational roots. The possible rational roots are factors of the constant term (24) divided by factors of the leading coefficient (1). The factors of 24 are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$.

3. Testing Possible Rational Roots: Let's substitute some of these values to see if they satisfy the equation. Through testing, it turns out that $x = 2$ is a root, because: $2^3 - 6(2^2) - 4(2) + 24 = 8 - 24 - 8 + 24 = 0$

4. Factoring the Polynomial: Since $x = 2$ is a root, we can factor $x - 2$ out of the polynomial. Using synthetic division: $x^3 - 6x^2 - 4x + 24 = (x - 2)(x^2 - 4x - 12)$

5. Factoring the Quadratic: Now, we need to factor $x^2 - 4x - 12$. The factors of -12 that add up to -4 are -6 and +2, so: $x^2 - 4x - 12 = (x - 6)(x + 2)$

6. Final Factorization: We now have: $x^3 - 6x^2 - 4x + 24 = (x - 2)(x - 6)(x + 2)$

7. Finding the Roots: Setting each factor equal to zero, we find: $x - 2 = 0 \Rightarrow x = 2$ $x - 6 = 0 \Rightarrow x = 6$ $x + 2 = 0 \Rightarrow x = -2$

8. Conclusion: The polynomial has 3 real roots: $x = 2$, $x = 6$, and $x = -2$.

Answer

The polynomial $x^3 - 6x^2 - 4x + 24 = 0$ has 3 solutions.

Would you like more details on any part of this solution, or have any questions?

Here are 5 follow-up questions related to this problem:

1. How would the number of solutions change if the polynomial had an even degree, such as $x^4$?
2. What methods can be used to solve polynomials that don’t have rational roots?
3. How does the Fundamental Theorem of Algebra determine the number of solutions?
4. What is synthetic division, and how does it help in factoring polynomials?
5. How can we determine if a polynomial has complex solutions?

Tip: When solving polynomial equations, checking for rational roots first can simplify the factoring process significantly.