Which of the following is true if `(x+1)` and `(x+2)` are factors of `p(x)=x^(3)+3x^(2)-2 alpha x+beta` ?

Options

A. `2alpha+2beta=2`
B. `2alpha-3beta=-2`
C. `alpha-7beta=5`
D. `7alpha-beta=2`

Answer

To solve the problem, we need to determine the relationship between the coefficients \( \alpha \) and \( \beta \) in the polynomial \( p(x) = x^3 + 3x^2 - 2\alpha x + \beta \) given that \( (x + 1) \) and \( (x + 2) \) are factors of the polynomial.

### Step-by-Step Solution:

1. **Use the Factor Theorem**: Since \( (x + 1) \) is a factor, substituting \( x = -1 \) into \( p(x) \) should yield 0.
\[ p(-1) = (-1)^3 + 3(-1)^2 - 2\alpha(-1) + \beta = 0 \]
Simplifying this gives:
\[ -1 + 3 + 2\alpha + \beta = 0 \]
\[ 2\alpha + \beta + 2 = 0 \quad \text{(Equation 1)} \]

2. **Substitute for the second factor**: Since \( (x + 2) \) is also a factor, substituting \( x = -2 \) into \( p(x) \) should also yield 0.
\[ p(-2) = (-2)^3 + 3(-2)^2 - 2\alpha(-2) + \beta = 0 \]
Simplifying this gives:
\[ -8 + 12 + 4\alpha + \beta = 0 \]
\[ 4\alpha + \beta + 4 = 0 \quad \text{(Equation 2)} \]

3. **Set up the system of equations**: Now we have two equations:
- \( 2\alpha + \beta + 2 = 0 \) (Equation 1)
- \( 4\alpha + \beta + 4 = 0 \) (Equation 2)

4. **Subtract the equations**: To eliminate \( \beta \), we can subtract Equation 1 from Equation 2:
\[ (4\alpha + \beta + 4) - (2\alpha + \beta + 2) = 0 \]
This simplifies to:
\[ 4\alpha - 2\alpha + 4 - 2 = 0 \]
\[ 2\alpha + 2 = 0 \]
\[ 2\alpha = -2 \quad \Rightarrow \quad \alpha = -1 \]

5. **Substitute back to find \( \beta \)**: Now substitute \( \alpha = -1 \) back into Equation 1:
\[ 2(-1) + \beta + 2 = 0 \]
\[ -2 + \beta + 2 = 0 \]
\[ \beta = 0 \]

6. **Conclusion**: We have found that \( \alpha = -1 \) and \( \beta = 0 \).

### Final Answer:
The values of \( \alpha \) and \( \beta \) are:
\[ \alpha = -1, \quad \beta = 0 \]