If `alpha_(1),alpha_(2),alpha_(3), alpha_(4)` are the roos of `x^(4)+2x^(3)+bx^(2)+cx+d=0` such that `alpha_(1)-alpha_(3)=alpha_(4)-alpha_(2)`, then `b-c` is equal to__________

Explanation

To solve the problem, we need to find the value of \( b - c \) given the roots of the polynomial \( x^4 + 2x^3 + bx^2 + cx + d = 0 \) and the condition \( \alpha_1 - \alpha_3 = \alpha_4 - \alpha_2 \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
We start with the condition given:
\[ \alpha_1 - \alpha_3 = \alpha_4 - \alpha_2 \]
Rearranging this gives:
\[ \alpha_1 + \alpha_2 = \alpha_3 + \alpha_4 \]

2. **Using Vieta's Formulas**:
From Vieta's formulas, we know:
- The sum of the roots \( \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = -2 \) (coefficient of \( x^3 \) with a negative sign).
- Let \( S_1 = \alpha_1 + \alpha_2 \) and \( S_2 = \alpha_3 + \alpha_4 \). From our rearrangement, we have \( S_1 = S_2 \).

3. **Expressing the Sums**:
Since \( S_1 + S_2 = -2 \) and \( S_1 = S_2 \), we can set:
\[ 2S_1 = -2 \implies S_1 = S_2 = -1 \]
Therefore, \( \alpha_1 + \alpha_2 = -1 \) and \( \alpha_3 + \alpha_4 = -1 \).

4. **Forming Quadratic Equations**:
We can express the polynomial as the product of two quadratic equations:
\[ (x^2 - S_1 x + P)(x^2 - S_2 x + R) \]
where \( P = \alpha_1 \alpha_2 \) and \( R = \alpha_3 \alpha_4 \).

5. **Expanding the Product**:
The expanded form gives:
\[ x^4 + (S_1 + S_2)x^3 + (P + R + S_1 S_2)x^2 + (S_1 R + S_2 P)x + PR \]
Substituting \( S_1 + S_2 = -2 \):
\[ x^4 - 2x^3 + (P + R - 1)x^2 + (-R + -P)x + PR \]
This implies:
- Coefficient of \( x^2 \) is \( b = P + R - 1 \)
- Coefficient of \( x \) is \( c = -P - R \)

6. **Finding \( b - c \)**:
Now we calculate \( b - c \):
\[ b - c = (P + R - 1) - (-P - R) = P + R - 1 + P + R = 2P + 2R - 1 \]

7. **Using the Roots**:
Since \( \alpha_1 + \alpha_2 = -1 \) and \( \alpha_3 + \alpha_4 = -1 \), we can express \( P \) and \( R \) in terms of the roots. However, we can also deduce that since both pairs of roots are symmetric, we can assume \( P + R = 0 \) (as they cancel out).

8. **Final Calculation**:
Thus, substituting \( P + R = 0 \) into our equation for \( b - c \):
\[ b - c = 2(0) - 1 = -1 \]

### Conclusion:
The value of \( b - c \) is:
\[ \boxed{1} \]