Yes, it is possible to factorize $x^3 + 8$ using the Factor Theorem.

The Factor Theorem states that if $f(a) = 0$, then $(x - a)$ is a factor of the polynomial $f(x)$.

Now, let's proceed with the factorization of $x^3 + 8$.

First, recognize that $x^3 + 8$ is a sum of cubes, because:

$x^3 + 8 = x^3 + 2^3$

The sum of cubes formula is:

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

In this case, $a = x$ and $b = 2$, so we can factor $x^3 + 8$ as:

$x^3 + 8 = (x + 2)(x^2 - 2x + 4)$

This is the factorized form of $x^3 + 8$.

Do you want further details on the Factor Theorem or the Sum of Cubes formula?

Here are some related questions you might find interesting:

1. How does the Factor Theorem help in identifying roots of a polynomial?
2. What is the difference between factoring a sum of cubes and factoring a difference of cubes?
3. How can we apply synthetic division to check if a factor is correct?
4. What would happen if the polynomial were $x^3 - 8$ instead?
5. How do we verify the factorization $(x + 2)(x^2 - 2x + 4)$?

Tip: Always look for recognizable patterns like the sum or difference of cubes when dealing with polynomials of degree 3.