Given that x+2 and x-3 are factors of `x^3+ax+b`, calculate the values 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 \( f(x) = x^3 + ax + b \).

### Step-by-Step Solution:

1. **Identify the roots from the factors**:
Since \( x + 2 \) and \( x - 3 \) are factors, we can set them to zero to find the roots:
\[ x + 2 = 0 \implies x = -2 \]
\[ x - 3 = 0 \implies x = 3 \]

2. **Evaluate the polynomial at the roots**:
Since both \( x = -2 \) and \( x = 3 \) are roots, we can substitute these values into the polynomial \( f(x) \) and set them equal to zero.

For \( x = -2 \):
\[ f(-2) = (-2)^3 + a(-2) + b = 0 \]
Simplifying this gives:
\[ -8 - 2a + b = 0 \quad \text{(Equation 1)} \]
Rearranging it:
\[ -2a + b = 8 \quad \text{(Equation 1)} \]

For \( x = 3 \):
\[ f(3) = 3^3 + a(3) + b = 0 \]
Simplifying this gives:
\[ 27 + 3a + b = 0 \quad \text{(Equation 2)} \]
Rearranging it:
\[ 3a + b = -27 \quad \text{(Equation 2)} \]

3. **Set up the system of equations**:
Now we have the following two equations:
\[ -2a + b = 8 \quad \text{(1)} \]
\[ 3a + b = -27 \quad \text{(2)} \]

4. **Eliminate \( b \)**:
We can eliminate \( b \) by subtracting Equation (1) from Equation (2):
\[ (3a + b) - (-2a + b) = -27 - 8 \]
This simplifies to:
\[ 3a + b + 2a - b = -35 \]
\[ 5a = -35 \]
Dividing both sides by 5 gives:
\[ a = -7 \]

5. **Substitute \( a \) back to find \( b \)**:
Now, substitute \( a = -7 \) back into Equation (1):
\[ -2(-7) + b = 8 \]
Simplifying gives:
\[ 14 + b = 8 \]
\[ b = 8 - 14 = -6 \]

6. **Final values**:
Thus, the values of \( a \) and \( b \) are:
\[ a = -7, \quad b = -6 \]

### Summary:
The values of \( a \) and \( b \) are:
\[ a = -7, \quad b = -6 \]