Find the quotient using long division. \[ \frac{6 x^{2}-14 x+1}{x-3} \]

To perform long division on the polynomial \( \frac{6x^2 - 14x + 1}{x - 3} \), we will follow the steps of polynomial long division. 1. **Divide the leading term**: Divide the leading term of the dividend \( 6x^2 \) by the leading term of the divisor \( x \): \[ \frac{6x^2}{x} = 6x \] This is the first term of the quotient. 2. **Multiply and subtract**: Multiply \( 6x \) by the entire divisor \( x - 3 \): \[ 6x(x - 3) = 6x^2 - 18x \] Now, subtract this from the original polynomial: \[ (6x^2 - 14x + 1) - (6x^2 - 18x) = -14x + 1 + 18x = 4x + 1 \] 3. **Repeat the process**: Now, we need to divide the new leading term \( 4x \) by the leading term of the divisor \( x \): \[ \frac{4x}{x} = 4 \] This is the next term of the quotient. 4. **Multiply and subtract again**: Multiply \( 4 \) by the entire divisor \( x - 3 \): \[ 4(x - 3) = 4x - 12 \] Now, subtract this from the current polynomial \( 4x + 1 \): \[ (4x + 1) - (4x - 12) = 1 + 12 = 13 \] 5. **Final result**: Since the degree of the remainder \( 13 \) is less than the degree of the divisor \( x - 3 \), we stop here. The quotient is \( 6x + 4 \) and the remainder is \( 13 \). Therefore, we can express the result of the division as: \[ \frac{6x^2 - 14x + 1}{x - 3} = 6x + 4 + \frac{13}{x - 3} \] Thus, the final answer is: \[ \boxed{6x + 4} \quad \text{with a remainder of } 13. \]