If the polynomial `f(x) =x^(3) + ax^(2) + bx +6` is divded by (x-3), the remainder is 3. Also, (x-2) is a factor of the polynimal f (x). Find the value of a and b.

Options

A. `a =-1, b =3`
B. `a= 1, b=-3`
C. `a=3, b=-1`
D. `a=-3, b=-1`

Answer

To solve the problem, we will follow these steps:

### Step 1: Use the Factor Theorem
Since \( (x - 2) \) is a factor of the polynomial \( f(x) \), we know that \( f(2) = 0 \).

Given the polynomial:
\[ f(x) = x^3 + ax^2 + bx + 6 \]
We substitute \( x = 2 \):
\[ f(2) = 2^3 + a(2^2) + b(2) + 6 = 0 \]
Calculating this gives:
\[ 8 + 4a + 2b + 6 = 0 \]
Simplifying:
\[ 4a + 2b + 14 = 0 \]
Thus, we can rewrite this as:
\[ 2a + b = -7 \quad \text{(Equation 1)} \]

### Step 2: Use the Remainder Theorem
We are also given that when \( f(x) \) is divided by \( (x - 3) \), the remainder is 3. This means \( f(3) = 3 \).

Substituting \( x = 3 \) into the polynomial:
\[ f(3) = 3^3 + a(3^2) + b(3) + 6 = 3 \]
Calculating this gives:
\[ 27 + 9a + 3b + 6 = 3 \]
Simplifying:
\[ 9a + 3b + 33 = 3 \]
Thus, we can rewrite this as:
\[ 9a + 3b = -30 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations
Now we have a system of equations:
1. \( 2a + b = -7 \) (Equation 1)
2. \( 9a + 3b = -30 \) (Equation 2)

We can simplify Equation 2 by dividing everything by 3:
\[ 3a + b = -10 \quad \text{(Equation 3)} \]

Now we can solve Equations 1 and 3 together:
From Equation 1:
\[ b = -7 - 2a \]
Substituting \( b \) into Equation 3:
\[ 3a + (-7 - 2a) = -10 \]
This simplifies to:
\[ 3a - 7 - 2a = -10 \]
\[ a - 7 = -10 \]
\[ a = -3 \]

### Step 4: Find the Value of \( b \)
Now that we have \( a \), we can substitute it back into Equation 1 to find \( b \):
\[ 2(-3) + b = -7 \]
\[ -6 + b = -7 \]
\[ b = -1 \]

### Final Values
Thus, the values of \( a \) and \( b \) are:
\[ a = -3, \quad b = -1 \]

### Summary of the Solution
- The values are \( a = -3 \) and \( b = -1 \).