Find the quotient using long division. \[ \frac{9 a^{3}-3 a^{2}-3 a+15}{3 a+2} \]

To perform long division on the polynomial \( \frac{9a^3 - 3a^2 - 3a + 15}{3a + 2} \), we will divide the numerator \( 9a^3 - 3a^2 - 3a + 15 \) by the denominator \( 3a + 2 \). 1. **Divide the leading term**: The leading term of the numerator is \( 9a^3 \) and the leading term of the denominator is \( 3a \). We divide these: \[ \frac{9a^3}{3a} = 3a^2 \] So, the first term of the quotient is \( 3a^2 \). 2. **Multiply and subtract**: Now, we multiply \( 3a^2 \) by the entire divisor \( 3a + 2 \): \[ 3a^2(3a + 2) = 9a^3 + 6a^2 \] Next, we subtract this from the original polynomial: \[ (9a^3 - 3a^2 - 3a + 15) - (9a^3 + 6a^2) = -3a^2 - 6a^2 - 3a + 15 = -9a^2 - 3a + 15 \] 3. **Repeat the process**: Now we divide the new leading term \( -9a^2 \) by the leading term of the divisor \( 3a \): \[ \frac{-9a^2}{3a} = -3a \] So, the next term of the quotient is \( -3a \). 4. **Multiply and subtract again**: We multiply \( -3a \) by the divisor \( 3a + 2 \): \[ -3a(3a + 2) = -9a^2 - 6a \] Now we subtract this from the current polynomial: \[ (-9a^2 - 3a + 15) - (-9a^2 - 6a) = -3a + 6a + 15 = 3a + 15 \] 5. **Continue the process**: Now we divide the leading term \( 3a \) by the leading term of the divisor \( 3a \): \[ \frac{3a}{3a} = 1 \] So, the next term of the quotient is \( 1 \). 6. **Multiply and subtract one last time**: We multiply \( 1 \) by the divisor \( 3a + 2 \): \[ 1(3a + 2) = 3a + 2 \] Now we subtract this from the current polynomial: \[ (3a + 15) - (3a + 2) = 15 - 2 = 13 \] 7. **Final result**: The division is complete, and we have a remainder of \( 13 \). Therefore, we can express the result as: \[ \frac{9a^3 - 3a^2 - 3a + 15}{3a + 2} = 3a^2 - 3a + 1 + \frac{13}{3a + 2} \] Thus, the final answer is: \[ \boxed{3a^2 - 3a + 1 + \frac{13}{3a + 2}} \]