Question

divide. fill in the quotient and the remainder.
\\((x^2 - 3x + 3) \div (x - 2) = \square + \frac{\square}{x - 2}\\)

Explanation:

Step1: Perform polynomial long division

Divide \(x^2 - 3x + 3\) by \(x - 2\). First, divide the leading term \(x^2\) by \(x\) to get \(x\). Multiply \(x - 2\) by \(x\) to get \(x^2 - 2x\). Subtract this from \(x^2 - 3x + 3\): \((x^2 - 3x + 3) - (x^2 - 2x) = -x + 3\).

Step2: Divide the new leading term

Now, divide \(-x\) by \(x\) to get \(-1\). Multiply \(x - 2\) by \(-1\) to get \(-x + 2\). Subtract this from \(-x + 3\): \((-x + 3) - (-x + 2) = 1\).

Step3: Write the result

The quotient is \(x - 1\) and the remainder is \(1\). So, \((x^2 - 3x + 3) \div (x - 2) = x - 1 + \frac{1}{x - 2}\).

Answer:

The quotient is \(x - 1\) and the remainder numerator is \(1\), so the first box is \(x - 1\) and the second box (remainder) is \(1\). Thus, \((x^2 - 3x + 3) \div (x - 2) = \boxed{x - 1} + \frac{\boxed{1}}{x - 2}\)