Find the quotient and remainder using long division. \( \frac{x^{3}-41 x+28}{x-6} \) If this is a quiz or exam question, all steps and logic must be shown for full credit. The quotient is The remainder is

Let’s divide \( x^3 - 41x + 28 \) by \( x - 6 \) using long division. 1. **Setup**: Write \( x^3 - 41x + 28 \) under the division symbol and \( x - 6 \) outside. 2. **First Division**: - Divide the leading term \( x^3 \) by \( x \) to get \( x^2 \). - Multiply \( x^2 \) by \( x - 6 \): \( x^2(x - 6) = x^3 - 6x^2 \). - Subtract the result from \( x^3 - 41x + 28 \): \[ (x^3 - 41x + 28) - (x^3 - 6x^2) = 6x^2 - 41x + 28 \] 3. **Second Division**: - Divide the leading term \( 6x^2 \) by \( x \) to get \( 6x \). - Multiply \( 6x \) by \( x - 6 \): \( 6x(x - 6) = 6x^2 - 36x \). - Subtract the result: \[ (6x^2 - 41x + 28) - (6x^2 - 36x) = -5x + 28 \] 4. **Third Division**: - Divide the leading term \( -5x \) by \( x \) to get \( -5 \). - Multiply \( -5 \) by \( x - 6 \): \( -5(x - 6) = -5x + 30 \). - Subtract the result: \[ (-5x + 28) - (-5x + 30) = -2 \] At this point, the division algorithm gives us: \[ \text{Quotient: } x^2 + 6x - 5 \] \[ \text{Remainder: } -2 \] Final Answer: The quotient is \( x^2 + 6x - 5 \) The remainder is \( -2 \)