Select the correct answer. Which expression is equivalent to \( \frac{(x-1)^{2}}{x^{2}-x-12} \cdot \frac{x^{2}+x-6}{x^{2}-6 x+5} \) if no denominator equals zero? A. \( \frac{x^{2}-3 x+2}{x^{2}-9 x+20} \) B. \( \frac{x^{2}+x-2}{x^{3}-x-20} \) C. \( \frac{x^{2}-3 x+2}{x^{2}-x-20} \) D. \( \frac{x^{2}+3 x+2}{x^{2}+x-20} \)

To simplify the expression \( \frac{(x-1)^{2}}{x^{2}-x-12} \cdot \frac{x^{2}+x-6}{x^{2}-6 x+5} \), we first factor all quadratics present: 1. Factor \( x^2 - x - 12 = (x - 4)(x + 3) \) 2. Factor \( x^2 + x - 6 = (x - 2)(x + 3) \) 3. Factor \( x^2 - 6x + 5 = (x - 5)(x - 1) \) Now substitute these factorizations into the original expression: \[ \frac{(x-1)^{2}}{(x - 4)(x + 3)} \cdot \frac{(x - 2)(x + 3)}{(x - 5)(x - 1)} \] Now, simplify this expression: 1. Cancel one \( (x - 1) \) term from the numerator and the denominator: \[ \frac{(x - 1) \cdot (x - 2)(x + 3)}{(x - 4)(x + 3)(x - 5)} \] 2. Cancel \( (x + 3) \) term: \[ \frac{(x - 1)(x - 2)}{(x - 4)(x - 5)} \] Now expand the numerator and denominator: - The numerator simplifies to \( x^2 - 3x + 2 \). - The denominator is already factored as \( (x - 4)(x - 5) = x^2 - 9x + 20 \). Putting it all together gives us: \[ \frac{x^2 - 3x + 2}{x^2 - 9x + 20} \] Since this matches option A, the correct answer is: **A. \( \frac{x^{2}-3 x+2}{x^{2}-9 x+20} \)**