To solve the problem, we need to find the values of $m$ and $n$ such that the polynomial $x^3 + 3x^2 - 2mx + n$ is divisible by both $x + 2$ and $x + 1$.

Step 1: Use the Factor Theorem

The Factor Theorem states that if $x + a$ is a factor of a polynomial, then the polynomial evaluated at $x = -a$ is equal to zero.

1. For $x + 2$ to be a factor, $f(-2) = 0$.
2. For $x + 1$ to be a factor, $f(-1) = 0$.

Let $f(x) = x^3 + 3x^2 - 2mx + n$.

Step 2: Apply the Factor Theorem

• Evaluate $f(-2)$: $f(-2) = (-2)^3 + 3(-2)^2 - 2m(-2) + n = -8 + 12 + 4m + n = 4m + n + 4 = 0$ This gives the equation: $4m + n + 4 = 0 \quad \text{(Equation 1)}$ Simplifying: $n = -4m - 4 \quad \text{(Equation 1)}$

• Evaluate $f(-1)$: $f(-1) = (-1)^3 + 3(-1)^2 - 2m(-1) + n = -1 + 3 + 2m + n = 2m + n + 2 = 0$ This gives the equation: $2m + n + 2 = 0 \quad \text{(Equation 2)}$ Simplifying: $n = -2m - 2 \quad \text{(Equation 2)}$

Step 3: Solve the System of Equations

We now have two equations:

1. $n = -4m - 4$
2. $n = -2m - 2$

Set the equations equal to each other: $-4m - 4 = -2m - 2$

Simplifying: $-2m = 2$ $m = -1$

Step 4: Substitute $m = -1$ back into one of the equations

Using Equation 2: $n = -2(-1) - 2 = 2 - 2 = 0$

Conclusion

The values are: $m = -1, \quad n = 0$

Would you like further details or have any questions? Here are some related questions to consider:

1. How would the solution change if different factors were provided?
2. Can the Factor Theorem be used for polynomials of any degree?
3. What if the polynomial had more than two factors?
4. How would you verify if your answers for $m$ and $n$ are correct?
5. Could you solve this problem graphically or using other methods?

Tip: Always check your solutions by substituting back into the original polynomial to ensure they satisfy the given conditions.