Let `alpha` be the real and `beta, gamma` be the complex roots of `x^(3) + 3x^(2) + 3x + 28 = 0`, then `|alpha + 2 beta + 2 gamma|` = _________________.

Answer

To solve the problem, we need to find the value of \( | \alpha + 2\beta + 2\gamma | \) where \( \alpha \) is the real root and \( \beta, \gamma \) are the complex roots of the polynomial \( x^3 + 3x^2 + 3x + 28 = 0 \).

### Step-by-step Solution:

1. **Identify the polynomial**: We have the polynomial \( f(x) = x^3 + 3x^2 + 3x + 28 \).

2. **Find the sum of the roots**: According to Vieta's formulas, the sum of the roots \( \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{3}{1} = -3 \).

3. **Determine the nature of the roots**: Since the polynomial is cubic and has a positive leading coefficient, we can analyze its behavior to find the roots. We will check for real roots by evaluating the polynomial at various points.

4. **Evaluate the polynomial at \( x = -4 \)**:
\[ f(-4) = (-4)^3 + 3(-4)^2 + 3(-4) + 28 = -64 + 48 - 12 + 28 = 0. \]
Thus, \( \alpha = -4 \) is a real root.

5. **Find \( \beta + \gamma \)**: Using the sum of the roots:
\[ \alpha + \beta + \gamma = -3 \implies -4 + \beta + \gamma = -3 \implies \beta + \gamma = 1. \]

6. **Calculate \( \alpha + 2\beta + 2\gamma \)**:
\[ \alpha + 2\beta + 2\gamma = -4 + 2(\beta + \gamma) = -4 + 2(1) = -4 + 2 = -2. \]

7. **Find the absolute value**:
\[ | \alpha + 2\beta + 2\gamma | = |-2| = 2. \]

### Final Answer:
\[ | \alpha + 2\beta + 2\gamma | = 2. \]