If (x + 2) and (x + 3) are factors of `x^(3)+ax+b`, find the value of 'a' and 'b'.

Answer

To solve the problem, we need to find the values of 'a' and 'b' given that \( (x + 2) \) and \( (x + 3) \) are factors of the polynomial \( x^3 + ax + b \).

### Step 1: Set up the equations using the factors
Since \( (x + 2) \) and \( (x + 3) \) are factors, we know that if we substitute \( x = -2 \) and \( x = -3 \) into the polynomial, the result should be zero.

### Step 2: Substitute \( x = -2 \)
Substituting \( x = -2 \) into the polynomial:
\[ (-2)^3 + a(-2) + b = 0 \]
This simplifies to:
\[ -8 - 2a + b = 0 \]
Rearranging gives us:
\[ -2a + b = 8 \quad \text{(Equation 1)} \]

### Step 3: Substitute \( x = -3 \)
Now, substituting \( x = -3 \) into the polynomial:
\[ (-3)^3 + a(-3) + b = 0 \]
This simplifies to:
\[ -27 - 3a + b = 0 \]
Rearranging gives us:
\[ -3a + b = 27 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we have a system of two equations:
1. \( -2a + b = 8 \)
2. \( -3a + b = 27 \)

We can eliminate \( b \) by subtracting Equation 1 from Equation 2:
\[ (-3a + b) - (-2a + b) = 27 - 8 \]
This simplifies to:
\[ -a = 19 \]
Thus, we find:
\[ a = -19 \]

### Step 5: Substitute back to find \( b \)
Now, substitute \( a = -19 \) back into Equation 1 to find \( b \):
\[ -2(-19) + b = 8 \]
This simplifies to:
\[ 38 + b = 8 \]
Rearranging gives:
\[ b = 8 - 38 = -30 \]

### Conclusion
The values of \( a \) and \( b \) are:
\[ a = -19, \quad b = -30 \]