Find the following quotient: \[ \frac{x^{4}-2 x^{3}+5 x^{2}-6 x+8}{x-1} \]

To find the quotient of the polynomial \( \frac{x^{4}-2 x^{3}+5 x^{2}-6 x+8}{x-1} \), we can perform polynomial long division. 1. **Set Up the Long Division:** We divide \( x^{4}-2 x^{3}+5 x^{2}-6 x+8 \) by \( x-1 \). 2. **Divide the Leading Terms:** - The leading term of the numerator is \( x^4 \) and the leading term of the denominator is \( x \). - Divide these to get \( x^3 \). 3. **Multiply and Subtract:** - Multiply \( x^3 \) by \( (x-1) \) to get \( x^{4}-x^{3} \). - Subtract this from \( x^{4}-2 x^{3}+5 x^{2}-6 x+8 \): \[ (x^{4}-2 x^{3}+5 x^{2}-6 x+8) - (x^{4}-x^{3}) = -x^{3}+5 x^{2}-6 x+8. \] 4. **Repeat the Process:** - Now take the new leading term, \( -x^3 \), and divide by \( x \): \(-x^3 \div x = -x^2\). - Multiply \( -x^2 \) by \( (x-1) \) to get \( -x^{3}+x^{2} \). - Subtract: \[ (-x^{3}+5 x^{2}-6 x+8) - (-x^{3}+x^{2}) = 4 x^{2}-6 x+8. \] 5. **Continue Dividing:** - Divide \( 4x^2 \) by \( x \) to get \( 4x \). - Multiply \( 4x \) by \( (x-1) \) to get \( 4x^{2}-4x \). - Subtract: \[ (4x^{2}-6x+8) - (4x^{2}-4x) = -2x+8. \] 6. **Final Steps:** - Now divide \( -2x \) by \( x \) to get \( -2 \). - Multiply \( -2 \) by \( (x-1) \) to get \( -2x+2 \). - Subtract: \[ (-2x+8) - (-2x+2) = 6. \] Putting it all together, we have: - Quotient: \( x^3 - x^2 + 4x - 2 \) - Remainder: \( 6 \) Thus, the result of the division is: \[ \frac{x^{4}-2 x^{3}+5 x^{2}-6 x+8}{x-1} = x^{3}-x^{2}+4x-2 + \frac{6}{x-1}. \] The final answer is: \[ \boxed{x^{3}-x^{2}+4x-2 + \frac{6}{x-1}}. \]