Find the values of m and n so that x-1 and x + 2 both are factors of `x^3+(3m+1)x^2+nx-18`.

Answer

To find the values of \( m \) and \( n \) such that \( x - 1 \) and \( x + 2 \) are factors of the polynomial \( f(x) = x^3 + (3m + 1)x^2 + nx - 18 \), we can follow these steps:

### Step 1: Set up the polynomial
Let \( f(x) = x^3 + (3m + 1)x^2 + nx - 18 \).

### Step 2: Use the Factor Theorem for \( x - 1 \)
Since \( x - 1 \) is a factor, we can substitute \( x = 1 \) into \( f(x) \) and set it equal to zero:
\[ f(1) = 1^3 + (3m + 1) \cdot 1^2 + n \cdot 1 - 18 = 0 \]
This simplifies to:
\[ 1 + (3m + 1) + n - 18 = 0 \]
\[ 3m + n - 16 = 0 \]
Thus, we have our first equation:
\[ 3m + n = 16 \quad \text{(Equation 1)} \]

### Step 3: Use the Factor Theorem for \( x + 2 \)
Since \( x + 2 \) is also a factor, we substitute \( x = -2 \) into \( f(x) \) and set it equal to zero:
\[ f(-2) = (-2)^3 + (3m + 1)(-2)^2 + n(-2) - 18 = 0 \]
This simplifies to:
\[ -8 + (3m + 1) \cdot 4 - 2n - 18 = 0 \]
\[ -8 + 12m + 4 - 2n - 18 = 0 \]
\[ 12m - 2n - 22 = 0 \]
Thus, we have our second equation:
\[ 12m - 2n = 22 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we have the system of equations:
1. \( 3m + n = 16 \)
2. \( 12m - 2n = 22 \)

From Equation 1, we can express \( n \) in terms of \( m \):
\[ n = 16 - 3m \]

Substituting this expression for \( n \) into Equation 2:
\[ 12m - 2(16 - 3m) = 22 \]
Expanding this gives:
\[ 12m - 32 + 6m = 22 \]
Combining like terms:
\[ 18m - 32 = 22 \]
Adding 32 to both sides:
\[ 18m = 54 \]
Dividing by 18:
\[ m = 3 \]

### Step 5: Find \( n \)
Now substituting \( m = 3 \) back into the expression for \( n \):
\[ n = 16 - 3(3) = 16 - 9 = 7 \]

### Final Answer
The values of \( m \) and \( n \) are:
\[ m = 3, \quad n = 7 \]