If `(x+1)` and `(x-2)` are factors of `x^(3)+ax^(2)-bx-6,` then find the values of a and b respectively.

Options

A. 2, 3
B. 3, 5
C. 5, 3
D. 2, 5

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) given that \( (x+1) \) and \( (x-2) \) are factors of the polynomial \( x^3 + ax^2 - bx - 6 \).

### Step-by-Step Solution:

1. **Use the Factor Theorem**: Since \( (x+1) \) is a factor, we can substitute \( x = -1 \) into the polynomial and set it equal to zero:
\[ f(-1) = (-1)^3 + a(-1)^2 - b(-1) - 6 = 0 \]
Simplifying this gives:
\[ -1 + a + b - 6 = 0 \]
\[ a + b - 7 = 0 \quad \text{(Equation 1)} \]

2. **Substitute for the second factor**: Now, since \( (x-2) \) is also a factor, we substitute \( x = 2 \) into the polynomial and set it equal to zero:
\[ f(2) = (2)^3 + a(2)^2 - b(2) - 6 = 0 \]
Simplifying this gives:
\[ 8 + 4a - 2b - 6 = 0 \]
\[ 4a - 2b + 2 = 0 \quad \text{(Equation 2)} \]

3. **Rearranging Equation 2**: We can rearrange Equation 2 to isolate \( b \):
\[ 4a - 2b = -2 \]
Dividing the entire equation by 2 gives:
\[ 2a - b = -1 \quad \text{(Equation 3)} \]

4. **Solve the system of equations**: Now we have two equations:
- Equation 1: \( a + b = 7 \)
- Equation 3: \( 2a - b = -1 \)

We can add these two equations to eliminate \( b \):
\[ (a + b) + (2a - b) = 7 - 1 \]
Simplifying gives:
\[ 3a = 6 \implies a = 2 \]

5. **Substituting back to find \( b \)**: Now that we have \( a = 2 \), we can substitute this value back into Equation 1:
\[ 2 + b = 7 \implies b = 5 \]

### Final Answer:
Thus, the values of \( a \) and \( b \) are:
\[ a = 2, \quad b = 5 \]