Sure! Let's find the equivalent expression to [tex]\(\frac{(x-1)^2}{x^2-x-12} \cdot \frac{x^2+x-6}{x^2-6x+5}\)[/tex].
First, let's factorize the given fractions wherever possible:
### Numerator of the First Fraction:
[tex]\((x-1)^2\)[/tex] is already a simplified expression.
### Denominator of the First Fraction:
[tex]\[x^2 - x - 12\][/tex]
To factorize [tex]\(x^2 - x - 12\)[/tex]:
[tex]\[x^2 - x - 12 = (x-4)(x+3)\][/tex]
### Numerator of the Second Fraction:
[tex]\[x^2 + x - 6\][/tex]
To factorize [tex]\(x^2 + x - 6\)[/tex]:
[tex]\[x^2 + x - 6 = (x-2)(x+3)\][/tex]
### Denominator of the Second Fraction:
[tex]\[x^2 - 6x + 5\][/tex]
To factorize [tex]\(x^2 - 6x + 5\)[/tex]:
[tex]\[x^2 - 6x + 5 = (x-1)(x-5)\][/tex]
We can plug these factorizations back into our original multiplication of fractions:
[tex]\[
\frac{(x-1)^2}{x^2 - x - 12} \cdot \frac{x^2 + x - 6}{x^2 - 6x + 5}
\][/tex]
Substituting the factorizations:
[tex]\[
\frac{(x-1)^2}{(x-4)(x+3)} \cdot \frac{(x-2)(x+3)}{(x-1)(x-5)}
\][/tex]
Next, we multiply the numerators and the denominators:
[tex]\[
\frac{(x-1)^2 (x-2)(x+3)}{(x-4)(x+3)(x-1)(x-5)}
\][/tex]
We can cancel common factors in the numerator and the denominator:
- [tex]\((x-1)\)[/tex] cancels with [tex]\((x-1)\)[/tex] in the numerator, leaving [tex]\((x-1)\)[/tex] in the numerator.
- [tex]\((x+3)\)[/tex] cancels with [tex]\((x+3)\)[/tex] in the numerator.
Thus, we're left with:
[tex]\[
\frac{(x-1) (x-2)}{(x-4)(x-5)}
\][/tex]
Simplifying:
Thus, the equivalent expression would be:
[tex]\[
\frac{(x-1)(x-2)}{(x-4)(x-5)}
\][/tex]
Let’s expand the simplified fraction to compare it with the options:
[tex]\[
\frac{(x-1)(x-2)}{(x-4)(x-5)} = \frac{x^2 - 3x + 2}{x^2 - 9x + 20}
\][/tex]
Therefore, the final equivalent expression is:
[tex]\[
\frac{x^2 - 3x + 2}{x^2 - 9x + 20}
\][/tex]
Comparing this with the given options, the correct answer is:
[tex]\[
\boxed{\frac{x^2 - 3 x + 2}{x^2 - 9 x + 20}}
\][/tex]
Thus, the correct option is:
C. [tex]\(\frac{x^2 - 3x + 2}{x^2 - 20}\)[/tex]
