The question asks which of the given options is a factor of the polynomial $x^3 - 5x^2 - 2x + 24$, knowing that $x - 3$ is one of its linear factors.

Steps to solve:

1. Given: One of the factors is $x - 3$, meaning that $x = 3$ is a root of the polynomial.
2. Polynomial Division: We can divide $x^3 - 5x^2 - 2x + 24$ by $x - 3$ using either synthetic or long division to find the quotient, which will help us identify the remaining factors.

Let's start by dividing $x^3 - 5x^2 - 2x + 24$ by $x - 3$.

Polynomial division of $x^3 - 5x^2 - 2x + 24$ by $x - 3$:

• Divide the leading term $x^3$ by $x$, which gives $x^2$.
• Multiply $x^2$ by $x - 3$, giving $x^3 - 3x^2$.
• Subtract $(x^3 - 3x^2)$ from $x^3 - 5x^2$, which results in $-2x^2$.
• Divide $-2x^2$ by $x$, which gives $-2x$.
• Multiply $-2x$ by $x - 3$, giving $-2x^2 + 6x$.
• Subtract $(-2x^2 + 6x)$ from $-2x^2 - 2x$, which results in $-8x$.
• Divide $-8x$ by $x$, which gives $-8$.
• Multiply $-8$ by $x - 3$, giving $-8x + 24$.
• Subtract $(-8x + 24)$ from $-8x + 24$, resulting in 0.

So, the quotient is $x^2 - 2x - 8$.

Factorization of $x^2 - 2x - 8$:

Now, we factor $x^2 - 2x - 8$. We need two numbers that multiply to give -8 and add to give -2. The numbers are 2 and -4.

Thus, $x^2 - 2x - 8$ factors as: $(x - 4)(x + 2)$

Conclusion:

The complete factorization of the polynomial $x^3 - 5x^2 - 2x + 24$ is: $(x - 3)(x - 4)(x + 2)$

Hence, one of the factors is $x + 2$, which corresponds to the second option.

Let me know if you need more details, and here are some related questions to expand on this:

1. How do you perform synthetic division instead of long division?
2. Can all cubic polynomials be factored using this method?
3. How can you verify that $x = 3$ is a root of the polynomial?
4. What happens if a polynomial has no real roots?
5. How can you use the Rational Root Theorem to find possible roots?

Tip: When factoring cubic polynomials, checking for simple roots like integers using synthetic division or the Rational Root Theorem can speed up the process.