Find the quotient and remainder using long division. \( \frac{x^{3}-4 x^{2}+7}{x-1} \) The quotient is

To solve \( \frac{x^{3}-4 x^{2}+7}{x-1} \) using long division, we follow these steps: 1. Divide the leading term of the dividend \(x^3\) by the leading term of the divisor \(x\) to get \(x^2\). 2. Multiply \(x^2\) by the entire divisor \((x - 1)\) to get \(x^3 - x^2\). 3. Subtract this from the original polynomial: \[ (x^{3} - 4x^{2} + 7) - (x^{3} - x^{2}) = -3x^{2} + 7. \] 4. Bring down the next term if necessary. Here, we bring down nothing, so we continue with \(-3x^2 + 7\). 5. Divide \(-3x^2\) by \(x\) to get \(-3x\). 6. Multiply \(-3x\) by \((x - 1)\) to get \(-3x^2 + 3x\). 7. Subtract this from the current polynomial: \[ (-3x^{2} + 7) - (-3x^{2} + 3x) = -3x + 7. \] 8. Now, divide \(-3x\) by \(x\) to get \(-3\). 9. Multiply \(-3\) by \((x - 1)\) to get \(-3x + 3\). 10. Subtract this from the current polynomial: \[ (-3x + 7) - (-3x + 3) = 4. \] The process concludes here, as the remainder is a constant and cannot be divided further by \(x - 1\). Therefore, the quotient is \(x^2 - 3x - 3\) and the remainder is \(4\). Quotient: \(x^2 - 3x - 3\) Remainder: \(4\)