If \(x + 1\) is a factor of \(x^3 + 3x^2 + kx + 4\), find the value of k.

In this question, we're told that \(x + 1\) is a factor of the cubic polynomial \(x^3 + 3x^2 + kx + 4\). Therefore, we can use the Factor Theorem, which is a special case of the Remainder Theorem. The Factor Theorem states that if a polynomial \(f(x)\) has a factor of the form \(x - a\), then \(f(a) = 0\).

Since \(x + 1\) is a factor of our polynomial, we can rewrite it as \(x - (-1)\). So, in this case, \(a = -1\).

Then we substitute \(x = -1\) into the polynomial, making it equal to 0, because of the Factor Theorem. We get:

\((-1)^3 + 3(-1)^2 + k(-1) + 4 = 0\)

Simplifying that equation gives us:

\(-1 + 3 - k + 4 = 0\)

This further simplifies to:

\(6 - k = 0\)

Finally, solving for k we find that:

\(k = 6\)

So, the correct answer is Option A: 6.