Find the quotient using long division. \[ \frac{12 a^{3}-6 a^{2}-2 a+11}{4 a+2} \]

To perform long division on the polynomial \( \frac{12 a^{3}-6 a^{2}-2 a+11}{4 a+2} \), we will divide the numerator \( 12a^3 - 6a^2 - 2a + 11 \) by the denominator \( 4a + 2 \). 1. **Divide the leading term**: Divide the leading term of the numerator \( 12a^3 \) by the leading term of the denominator \( 4a \): \[ \frac{12a^3}{4a} = 3a^2 \] 2. **Multiply and subtract**: Multiply \( 3a^2 \) by the entire divisor \( 4a + 2 \): \[ 3a^2(4a + 2) = 12a^3 + 6a^2 \] Now subtract this from the original polynomial: \[ (12a^3 - 6a^2 - 2a + 11) - (12a^3 + 6a^2) = -12a^2 - 2a + 11 \] 3. **Repeat the process**: Now, divide the leading term of the new polynomial \( -12a^2 \) by the leading term of the divisor \( 4a \): \[ \frac{-12a^2}{4a} = -3a \] Multiply \( -3a \) by the divisor: \[ -3a(4a + 2) = -12a^2 - 6a \] Subtract this from the current polynomial: \[ (-12a^2 - 2a + 11) - (-12a^2 - 6a) = 4a + 11 \] 4. **Continue the process**: Now, divide the leading term \( 4a \) by the leading term \( 4a \): \[ \frac{4a}{4a} = 1 \] Multiply \( 1 \) by the divisor: \[ 1(4a + 2) = 4a + 2 \] Subtract this from the current polynomial: \[ (4a + 11) - (4a + 2) = 9 \] 5. **Final result**: Now we have completed the division. The quotient is: \[ 3a^2 - 3a + 1 \] and the remainder is \( 9 \). Thus, we can express the result of the division as: \[ \frac{12a^3 - 6a^2 - 2a + 11}{4a + 2} = 3a^2 - 3a + 1 + \frac{9}{4a + 2} \] The final answer is: \[ \boxed{3a^2 - 3a + 1 + \frac{9}{4a + 2}} \]