Question

question 3 (multiple choice worth 6 points) (03.01r mc) evaluate \\(\frac{x^3 - 2x^2 + 3x - 7}{x - 1}\\). \\(x^2 - x + 2 - \frac{5}{x - 1}\\) \\(x^2 - 2x + 2 - \frac{9}{x - 1}\\) \\(x^2 - x + 4 - \frac{9}{x - 1}\\) \\(x^3 - 2x + 2 - \frac{5}{x - 1}\\)

Explanation:

Step1: Perform Polynomial Long Division

Divide \(x^3 - 2x^2 + 3x - 7\) by \(x - 1\).

• Divide the leading term \(x^3\) by \(x\) to get \(x^2\). Multiply \(x - 1\) by \(x^2\) to get \(x^3 - x^2\). Subtract this from the original polynomial: \((x^3 - 2x^2 + 3x - 7) - (x^3 - x^2) = -x^2 + 3x - 7\).

Step2: Continue the Division

Divide the new leading term \(-x^2\) by \(x\) to get \(-x\). Multiply \(x - 1\) by \(-x\) to get \(-x^2 + x\). Subtract this from \(-x^2 + 3x - 7\): \((-x^2 + 3x - 7) - (-x^2 + x) = 2x - 7\).

Step3: Continue the Division

Divide the new leading term \(2x\) by \(x\) to get \(2\). Multiply \(x - 1\) by \(2\) to get \(2x - 2\). Subtract this from \(2x - 7\): \((2x - 7) - (2x - 2) = -5\).

Step4: Write the Result

The division gives \(x^2 - x + 2\) with a remainder of \(-5\). So, \(\frac{x^3 - 2x^2 + 3x - 7}{x - 1} = x^2 - x + 2 - \frac{5}{x - 1}\).

Answer:

A. \(x^2 - x + 2 - \frac{5}{x - 1}\)