If `x^3+ ax^2+ bx + 6` has x -2 as a factor and leaves a remainder 3 when divided by x- 3, find the values of a and b.

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) given the polynomial \( f(x) = x^3 + ax^2 + bx + 6 \). We know that \( x - 2 \) is a factor of this polynomial, and it leaves a remainder of 3 when divided by \( x - 3 \).

### Step 1: Use the Factor Theorem
Since \( x - 2 \) is a factor, by the Factor Theorem, we have:
\[ f(2) = 0 \]

Substituting \( x = 2 \) into the polynomial:
\[ f(2) = 2^3 + a(2^2) + b(2) + 6 = 0 \]
This simplifies to:
\[ 8 + 4a + 2b + 6 = 0 \]
Combining like terms gives:
\[ 4a + 2b + 14 = 0 \]
Thus, we can rewrite this as:
\[ 4a + 2b = -14 \quad \text{(Equation 1)} \]

### Step 2: Use the Remainder Theorem
Since the polynomial leaves a remainder of 3 when divided by \( x - 3 \), we have:
\[ f(3) = 3 \]

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

### Step 3: Solve the System of Equations
Now we have two equations:
1. \( 4a + 2b = -14 \)
2. \( 9a + 3b = -30 \)

To eliminate \( b \), we can multiply Equation 1 by 3 and Equation 2 by 2:
\[ 3(4a + 2b) = 3(-14) \implies 12a + 6b = -42 \quad \text{(Equation 3)} \]
\[ 2(9a + 3b) = 2(-30) \implies 18a + 6b = -60 \quad \text{(Equation 4)} \]

### Step 4: Subtract the Equations
Now we subtract Equation 3 from Equation 4:
\[ (18a + 6b) - (12a + 6b) = -60 + 42 \]
This simplifies to:
\[ 6a = -18 \]
Thus, we find:
\[ a = -3 \]

### Step 5: Substitute \( a \) Back to Find \( b \)
Now we substitute \( a = -3 \) back into Equation 1:
\[ 4(-3) + 2b = -14 \]
This simplifies to:
\[ -12 + 2b = -14 \]
Adding 12 to both sides gives:
\[ 2b = -2 \]
Thus, we find:
\[ b = -1 \]

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