If `x + 3` is a factor of `x^(3) + ax^(2) - bx + 6` and a + b = 7. Then, the values of a and b are respectively

Options

A. `1,6`
B. `0,7`
C. `0,-7`
D. `2,5`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We are given that \( x + 3 \) is a factor of the polynomial \( x^3 + ax^2 - bx + 6 \). According to the factor theorem, if \( x + 3 \) is a factor, then substituting \( x = -3 \) into the polynomial should yield zero.

### Step 2: Substitute \( x = -3 \)
We substitute \( x = -3 \) into the polynomial:

\[ f(-3) = (-3)^3 + a(-3)^2 - b(-3) + 6 \]

Calculating each term:

\[ = -27 + 9a + 3b + 6 \]

### Step 3: Set the equation to zero
Since \( x + 3 \) is a factor, we set \( f(-3) = 0 \):

\[ -27 + 9a + 3b + 6 = 0 \]

Simplifying this:

\[ 9a + 3b - 21 = 0 \]

### Step 4: Rearranging the equation
We can rearrange this equation:

\[ 9a + 3b = 21 \]

Dividing through by 3 gives us:

\[ 3a + b = 7 \quad \text{(Equation 1)} \]

### Step 5: Use the second condition
We are also given that \( a + b = 7 \) (Equation 2).

### Step 6: Solve the system of equations
Now we have two equations:

1. \( 3a + b = 7 \)
2. \( a + b = 7 \)

We can subtract Equation 2 from Equation 1:

\[ (3a + b) - (a + b) = 7 - 7 \]

This simplifies to:

\[ 2a = 0 \]

### Step 7: Solve for \( a \)
Dividing by 2 gives:

\[ a = 0 \]

### Step 8: Substitute \( a \) back to find \( b \)
Now we substitute \( a = 0 \) back into Equation 2:

\[ 0 + b = 7 \]

Thus, we find:

\[ b = 7 \]

### Final Answer
The values of \( a \) and \( b \) are:

\[ a = 0, \quad b = 7 \]