To solve the given problem and find an equivalent expression for [tex]\(\frac{(x-1)^2}{x^2-x-12} \cdot \frac{x^2+x-6}{x^2-6x+5}\)[/tex], let's go through the steps in detail.
### Step 1: Factorize the Denominators
First, factorize the denominators of the expressions:
- [tex]\(x^2 - x - 12\)[/tex] can be factorized as [tex]\((x - 4)(x + 3)\)[/tex]:
[tex]\[x^2 - x - 12 = (x - 4)(x + 3)\][/tex]
- [tex]\(x^2 - 6x + 5\)[/tex] can be factorized as [tex]\((x - 1)(x - 5)\)[/tex]:
[tex]\[x^2 - 6x + 5 = (x - 1)(x - 5)\][/tex]
### Step 2: Substitute the Factored Forms
Substitute the factored forms into the rational expressions:
[tex]\[
\frac{(x-1)^2}{x^2-x-12} \cdot \frac{x^2 + x - 6}{x^2 - 6x + 5} = \frac{(x-1)^2}{(x-4)(x+3)} \cdot \frac{x^2 + x - 6}{(x-1)(x-5)}
\][/tex]
### Step 3: Factor the Numerators
Next, we factorize the numerators if possible:
- [tex]\((x - 1)^2\)[/tex] is already factored.
- [tex]\(x^2 + x - 6\)[/tex] can be factorized as [tex]\((x - 2)(x + 3)\)[/tex]:
[tex]\[x^2 + x - 6 = (x - 2)(x + 3)\][/tex]
### Step 4: Substitute the Factored Numerators
Now, substitute the factored numerators back into the expression:
[tex]\[
\frac{(x-1)^2}{(x-4)(x+3)} \cdot \frac{(x-2)(x+3)}{(x-1)(x-5)}
\][/tex]
### Step 5: Simplify the Expression
Combine the expressions and cancel out the common factors:
[tex]\[
\frac{(x-1)^2 (x-2)(x+3)}{(x-4)(x+3)(x-1)(x-5)}
\][/tex]
Notice that [tex]\((x-1)\)[/tex] and [tex]\((x+3)\)[/tex] are common factors in the numerator and denominator:
[tex]\[
= \frac{(x-1) (x-2)}{(x-4)(x-5)}
\][/tex]
Therefore, the simplified form of the rational expression is:
[tex]\[
\frac{(x-2)(x-1)}{(x-4)(x-5)}
\][/tex]
Match this final expression with the options given:
[tex]\[
\frac{(x-2)(x-1)}{(x-4)(x-5)} = \frac{(x - 2)(x - 1)}{(x - 4)(x - 5)}
\][/tex]
Which corresponds to Option A, since:
[tex]\[
\boxed{\frac{x^2-3 x+2}{x^2-9 x+20}}
\][/tex]
is the same as:
[tex]\[
\frac{(x - 2)(x - 1)}{(x - 4)(x - 5)}
\][/tex]
Thus, the correct answer is:
A. [tex]\(\frac{x^2-3 x+2}{x^2-9 x+20}\)[/tex]
