Since the polynomial $ax^2 + bx + c$ is divisible by $(x + 2)$ and $(x + 3)$, we know that $x = -2$ and $x = -3$ are roots of the polynomial. Therefore:

1. $f(-2) = 0$
2. $f(-3) = 0$

Additionally, we are given that dividing the polynomial by $(x - 1)$ leaves a remainder of 36, which implies:

3. $f(1) = 36$

Let's use these conditions to form equations based on the polynomial $f(x) = ax^2 + bx + c$.

Step 1: Set Up the Equations

Since $f(x) = ax^2 + bx + c$:

1. Using $f(-2) = 0$: $a(-2)^2 + b(-2) + c = 0$ $4a - 2b + c = 0$

2. Using $f(-3) = 0$: $a(-3)^2 + b(-3) + c = 0$ $9a - 3b + c = 0$

3. Using $f(1) = 36$: $a(1)^2 + b(1) + c = 36$ $a + b + c = 36$

Now we have a system of three equations:

$4a - 2b + c = 0$ $9a - 3b + c = 0$ $a + b + c = 36$

Step 2: Solve the System of Equations

1. Subtract the first equation from the second: $(9a - 3b + c) - (4a - 2b + c) = 0$ $5a - b = 0$ $b = 5a$

2. Substitute $b = 5a$ into the third equation: $a + (5a) + c = 36$ $6a + c = 36$ $c = 36 - 6a$

3. Substitute $b = 5a$ and $c = 36 - 6a$ into the first equation: $4a - 2(5a) + (36 - 6a) = 0$ $4a - 10a + 36 - 6a = 0$ $-12a + 36 = 0$ $-12a = -36$ $a = 3$

4. Substitute $a = 3$ back to find $b$ and $c$: $b = 5a = 5 \times 3 = 15$ $c = 36 - 6a = 36 - 6 \times 3 = 36 - 18 = 18$

Conclusion

The value of $c$ is $\boxed{18}$.

Would you like further details on this solution, or do you have any questions?

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Here are some related questions to consider:

1. What if the remainder when divided by $(x - 1)$ was different? How would that change the solution?
2. How can you verify the values of $a$, $b$, and $c$ once found?
3. What are some other methods to solve similar polynomial problems?
4. How does polynomial division affect the remainder theorem in these scenarios?
5. How would the solution change if additional roots were specified?

Tip: When solving systems of equations, isolating one variable can simplify the process significantly, especially in substitution methods.