If $(x+2)$ is a factor of the polynomial $4x^3 - 3x^2 + 2a + 6$, then substituting $x = -2$ should make the polynomial equal to zero. This is because any factor of a polynomial will satisfy $f(c) = 0$, where $c$ is the root.

Let's set up the polynomial function as:

$f(x) = 4x^3 - 3x^2 + 2a + 6.$

Now, substitute $x = -2$ and set $f(-2) = 0$:

$f(-2) = 4(-2)^3 - 3(-2)^2 + 2a + 6 = 0.$

Calculate each term:

$4(-2)^3 = 4(-8) = -32,$ $-3(-2)^2 = -3(4) = -12,$ $2a = 2a,$ $6 = 6.$

Substitute these values:

$-32 - 12 + 2a + 6 = 0.$

Simplify:

$-38 + 2a = 0.$

Solve for $a$:

$2a = 38,$ $a = 19.$

Therefore, the value of $a$ is:

$\boxed{19}.$

Let me know if you'd like a detailed explanation or have any questions!

Relative Questions

1. How do you check if a polynomial is divisible by a linear factor?
2. What is the Remainder Theorem and how is it used in factorization?
3. Can a polynomial have more than one linear factor?
4. What are synthetic division and long division methods for polynomials?
5. How do you find the roots of a cubic polynomial?

Tip: When checking if a polynomial has a particular linear factor, always substitute the root of the factor into the polynomial and verify if it results in zero.