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

Answer

To find the value of \( a \) such that the division of \( ax^3 + 9x^2 + 4x - 10 \) by \( x + 3 \) leaves a remainder of 5, we can follow these steps:

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

### Step 2: Identify the divisor
The divisor is \( x + 3 \). To find the value of \( x \) that makes the divisor zero, we set:
\[ x + 3 = 0 \implies x = -3 \]

### Step 3: Use the Remainder Theorem
According to the Remainder Theorem, the remainder when \( f(x) \) is divided by \( x + 3 \) is \( f(-3) \). We are given that this remainder is 5:
\[ f(-3) = 5 \]

### Step 4: Substitute \( -3 \) into the polynomial
Now we substitute \( -3 \) into \( f(x) \):
\[ f(-3) = a(-3)^3 + 9(-3)^2 + 4(-3) - 10 \]

### Step 5: Simplify the expression
Calculating each term:
- \( (-3)^3 = -27 \)
- \( 9(-3)^2 = 9 \times 9 = 81 \)
- \( 4(-3) = -12 \)

Now substituting these values:
\[ f(-3) = a(-27) + 81 - 12 - 10 \]
\[ f(-3) = -27a + 81 - 12 - 10 \]
\[ f(-3) = -27a + 81 - 22 \]
\[ f(-3) = -27a + 59 \]

### Step 6: Set the equation equal to the remainder
Since we know \( f(-3) = 5 \), we set up the equation:
\[ -27a + 59 = 5 \]

### Step 7: Solve for \( a \)
Now, we solve for \( a \):
\[ -27a + 59 = 5 \]
Subtract 59 from both sides:
\[ -27a = 5 - 59 \]
\[ -27a = -54 \]
Now, divide by -27:
\[ a = \frac{-54}{-27} = 2 \]

### Conclusion
Thus, the value of \( a \) is \( 2 \).

---