Question

question 18 of 40
perform the indicated operation.
\\((3x^2 - 6x) \div \dfrac{x^2 - 2x}{x^2 - 3x + 2}\\)
\\((3x^2 - 6x) \div \dfrac{x^2 - 2x}{x^2 - 3x + 2} =\\)

Explanation:

Step1: Factor each expression

Factor \(3x^2 - 6x\): \(3x(x - 2)\)
Factor \(x^2 - 2x\): \(x(x - 2)\)
Factor \(x^2 - 3x + 2\): \((x - 1)(x - 2)\)

Step2: Rewrite division as multiplication

\((3x^2 - 6x) \div \frac{x^2 - 2x}{x^2 - 3x + 2} = (3x^2 - 6x) \times \frac{x^2 - 3x + 2}{x^2 - 2x}\)
Substitute factored forms: \(3x(x - 2) \times \frac{(x - 1)(x - 2)}{x(x - 2)}\)

Step3: Cancel common factors

Cancel \(x\) ( \(x
eq0\) ), cancel \((x - 2)\) ( \(x
eq2\) ): \(3(x - 2)(x - 1)\)
Expand: \(3(x^2 - 3x + 2) = 3x^2 - 9x + 6\) (or leave as \(3(x - 1)(x - 2)\) )

Answer:

\(3(x - 1)(x - 2)\) (or \(3x^2 - 9x + 6\))