Find `alpha` and `beta` if x+1 and x+2 are factors of `p(x)=x^(3)+3x^(2)-2alphax+beta`.

Answer

To find the values of `alpha` and `beta` given that \( x + 1 \) and \( x + 2 \) are factors of the polynomial \( p(x) = x^3 + 3x^2 - 2\alpha x + \beta \), we can follow these steps:

### Step 1: Set up the equations using the factors
Since \( x + 1 \) and \( x + 2 \) are factors, we know that:
- \( p(-1) = 0 \)
- \( p(-2) = 0 \)

### Step 2: Calculate \( p(-1) \)
Substituting \( x = -1 \) into the polynomial:
\[ p(-1) = (-1)^3 + 3(-1)^2 - 2\alpha(-1) + \beta \]
Calculating this gives:
\[ p(-1) = -1 + 3 + 2\alpha + \beta = 2 + 2\alpha + \beta \]
Setting this equal to zero (since \( p(-1) = 0 \)):
\[ 2 + 2\alpha + \beta = 0 \quad \text{(Equation 1)} \]

### Step 3: Calculate \( p(-2) \)
Now substituting \( x = -2 \) into the polynomial:
\[ p(-2) = (-2)^3 + 3(-2)^2 - 2\alpha(-2) + \beta \]
Calculating this gives:
\[ p(-2) = -8 + 3(4) + 4\alpha + \beta = -8 + 12 + 4\alpha + \beta = 4 + 4\alpha + \beta \]
Setting this equal to zero (since \( p(-2) = 0 \)):
\[ 4 + 4\alpha + \beta = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( 2 + 2\alpha + \beta = 0 \)
2. \( 4 + 4\alpha + \beta = 0 \)

We can solve these equations simultaneously.

From Equation 1, we can express \( \beta \):
\[ \beta = -2 - 2\alpha \quad \text{(Substituting into Equation 2)} \]

Substituting \( \beta \) into Equation 2:
\[ 4 + 4\alpha + (-2 - 2\alpha) = 0 \]
Simplifying this:
\[ 4 + 4\alpha - 2 - 2\alpha = 0 \]
\[ 2 + 2\alpha = 0 \]
\[ 2\alpha = -2 \]
\[ \alpha = -1 \]

### Step 5: Substitute back to find \( \beta \)
Now substituting \( \alpha = -1 \) back into Equation 1:
\[ 2 + 2(-1) + \beta = 0 \]
\[ 2 - 2 + \beta = 0 \]
\[ \beta = 0 \]

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