Find the value of 'a' if the division of `ax^(3)+9x^(2)+4x-10` by x+3 leaves a remainder of 5.

Answer

To solve the problem, we need to find the value of 'a' such that the polynomial \( ax^3 + 9x^2 + 4x - 10 \) when divided by \( x + 3 \) leaves a remainder of 5. We can use the Remainder Theorem for this purpose.

### Step-by-Step Solution:

1. **Identify the polynomial**:
Let \( f(x) = ax^3 + 9x^2 + 4x - 10 \).

2. **Apply the Remainder Theorem**:
According to the Remainder Theorem, the remainder of the division of \( f(x) \) by \( x + 3 \) can be found by evaluating \( f(-3) \). We know that this remainder should equal 5.

3. **Set up the equation**:
We need to find \( f(-3) \):
\[ f(-3) = a(-3)^3 + 9(-3)^2 + 4(-3) - 10 \]
Simplifying this gives:
\[ f(-3) = a(-27) + 9(9) - 12 - 10 \]
\[ f(-3) = -27a + 81 - 12 - 10 \]
\[ f(-3) = -27a + 81 - 22 \]
\[ f(-3) = -27a + 59 \]

4. **Set the equation equal to the remainder**:
Since we want the remainder to be 5, we set up the equation:
\[ -27a + 59 = 5 \]

5. **Solve for 'a'**:
Rearranging the equation:
\[ -27a = 5 - 59 \]
\[ -27a = -54 \]
Dividing both sides by -27:
\[ a = \frac{-54}{-27} = 2 \]

6. **Final answer**:
Thus, the value of \( a \) is \( 2 \).