The minimum value of $ x(x \plus{} 4)(x \plus{} 8)(x \plus{} 12)$ in real numbers is ?

$\textbf{(A)}\  \minus{} 240 \qquad\textbf{(B)}\  \minus{} 252 \qquad\textbf{(C)}\  \minus{} 256 \qquad\textbf{(D)}\  \minus{} 260 \qquad\textbf{(E)}\  \minus{} 280$
1. **Substitution**: Let \( x = y - 8 \). Then the expression \( x(x + 4)(x + 8)(x + 12) \) becomes:
   \[
   (y - 8)(y - 4)y(y + 4)
   \]

2. **Rewriting the expression**: Notice that the expression can be grouped and rewritten using the difference of squares:
   \[
   (y - 8)(y - 4)y(y + 4) = [(y - 4) - 4][(y - 4) + 4]y(y + 4)
   \]
   Simplifying further:
   \[
   (y^2 - 16)(y^2 - 4y)
   \]

3. **Using the difference of squares**: We can rewrite the product of these two terms:
   \[
   (y^2 - 16)(y^2 - 4y) = (y^2 - 4y - 16)^2 - 16^2
   \]

4. **Finding the minimum value**: To find the minimum value of the expression, we need to analyze the term \( (y^2 - 4y - 16)^2 - 16^2 \). The term \( (y^2 - 4y - 16)^2 \) is always non-negative since it is a square. The minimum value of \( (y^2 - 4y - 16)^2 \) is 0, which occurs when:
   \[
   y^2 - 4y - 16 = 0
   \]
   Solving this quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   y = \frac{4 \pm \sqrt{16 + 64}}{2} = \frac{4 \pm \sqrt{80}}{2} = \frac{4 \pm 4\sqrt{5}}{2} = 2 \pm 2\sqrt{5}
   \]

5. **Evaluating the minimum value**: When \( y = 2 \pm 2\sqrt{5} \), the term \( (y^2 - 4y - 16)^2 \) becomes 0. Therefore, the minimum value of the original expression is:
   \[
   (0)^2 - 16^2 = -256
   \]

The final answer is \(\boxed{-256}\).