Given that x -2 and x + 1 are factors of `f(x)=x^(3)+3x^(2)+ax+b`, calculate the values of a and b. Hence, find all the factors of f(x).

Answer

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

### Step 1: Use the Factor Theorem
Since \( x - 2 \) is a factor, we can substitute \( x = 2 \) into \( f(x) \) and set it equal to zero:

\[ f(2) = 2^3 + 3(2^2) + a(2) + b = 0 \]

Calculating this gives:

\[ f(2) = 8 + 3(4) + 2a + b = 0 \]
\[ 8 + 12 + 2a + b = 0 \]
\[ 20 + 2a + b = 0 \]

This simplifies to:

\[ 2a + b = -20 \quad \text{(Equation 1)} \]

### Step 2: Use the Factor Theorem Again
Now, since \( x + 1 \) is also a factor, we substitute \( x = -1 \) into \( f(x) \):

\[ f(-1) = (-1)^3 + 3(-1)^2 + a(-1) + b = 0 \]

Calculating this gives:

\[ f(-1) = -1 + 3(1) - a + b = 0 \]
\[ -1 + 3 - a + b = 0 \]
\[ 2 - a + b = 0 \]

This simplifies to:

\[ -a + b = -2 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations
Now we have a system of equations:

1. \( 2a + b = -20 \) (Equation 1)
2. \( -a + b = -2 \) (Equation 2)

We can eliminate \( b \) by subtracting Equation 2 from Equation 1:

\[ (2a + b) - (-a + b) = -20 - (-2) \]
\[ 2a + b + a - b = -20 + 2 \]
\[ 3a = -18 \]
\[ a = -6 \]

### Step 4: Substitute \( a \) Back to Find \( b \)
Now substitute \( a = -6 \) back into Equation 2:

\[ -a + b = -2 \]
\[ -(-6) + b = -2 \]
\[ 6 + b = -2 \]
\[ b = -2 - 6 = -8 \]

### Step 5: Write the Polynomial with Found Values
Now we have \( a = -6 \) and \( b = -8 \). Thus, the polynomial becomes:

\[ f(x) = x^3 + 3x^2 - 6x - 8 \]

### Step 6: Factor the Polynomial
We already know two factors: \( x - 2 \) and \( x + 1 \). To find the third factor, we can perform polynomial long division of \( f(x) \) by \( (x - 2)(x + 1) \).

First, calculate \( (x - 2)(x + 1) \):

\[ (x - 2)(x + 1) = x^2 - 2x + x - 2 = x^2 - x - 2 \]

Now divide \( f(x) \) by \( x^2 - x - 2 \):

1. Divide the leading term: \( x^3 \div x^2 = x \)
2. Multiply \( x \) by \( x^2 - x - 2 \): \( x^3 - x^2 - 2x \)
3. Subtract from \( f(x) \):

\[ (x^3 + 3x^2 - 6x - 8) - (x^3 - x^2 - 2x) = 4x^2 - 4x - 8 \]

4. Divide the leading term: \( 4x^2 \div x^2 = 4 \)
5. Multiply \( 4 \) by \( x^2 - x - 2 \): \( 4x^2 - 4x - 8 \)
6. Subtract:

\[ (4x^2 - 4x - 8) - (4x^2 - 4x - 8) = 0 \]

Thus, the complete factorization of \( f(x) \) is:

\[ f(x) = (x - 2)(x + 1)(x + 4) \]

### Final Answer
The values of \( a \) and \( b \) are \( a = -6 \) and \( b = -8 \). The complete factorization of \( f(x) \) is:

\[ f(x) = (x - 2)(x + 1)(x + 4) \]