Which of the following properly factors the expression to prepare for reducing?
(A) $\frac{x^{2}-3 x+2}{(x-2)^{2}}=\frac{(x-3)(x-1)}{(x-2)(x-2)}$
(B) $\frac{x^{2}-3 x+2}{(x-2)^{2}}=\frac{(x-3)(x-1)}{(x-2)(x+2)}$
(C) $\frac{x^{2}-3 x+2}{(x-2)^{2}}=\frac{(x-1)(x-2)}{(x-2)(x-2)}$
(D) $\frac{x^{2}-3 x+2}{(x-2)^{2}}=\frac{(x-1)(x-2)}{(x-2)(x+2)}$

Answer

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Answer

Final Answer: $\boxed{(C)}$

Steps

Step 1 ：The problem is asking to factorize the expression $ \frac{x^{2}-3 x+2}{(x-2)^{2}} $.

Step 2 ：The numerator $ x^{2}-3 x+2 $ can be factorized into $ (x-1)(x-2) $.

Step 3 ：The denominator is already in the factorized form $ (x-2)^{2} $.

Step 4 ：So, the correct factorization of the given expression is $ \frac{(x-1)(x-2)}{(x-2)(x-2)} $.

Step 5 ：Checking the options, we find that this matches with option (C).

Step 6 ：Final Answer: $\boxed{(C)}$