1. **State the problem:** Simplify the expression $$\frac{x^3 - 7x + 6}{x^2 + 4x - 5} \times \frac{x^2 + 9x + 20}{x^3 - 4x^2 + 5x - 2}$$. 2. **Factor each polynomial:** - Numerator of first fraction: $$x^3 - 7x + 6$$. Try to find roots by inspection or Rational Root Theorem. Test $x=1$: $1^3 - 7(1) + 6 = 1 - 7 + 6 = 0$, so $(x-1)$ is a factor. Divide $x^3 - 7x + 6$ by $(x-1)$: $$\frac{x^3 - 7x + 6}{x-1} = x^2 + x - 6$$. Factor $x^2 + x - 6$: $$x^2 + x - 6 = (x+3)(x-2)$$. So, $$x^3 - 7x + 6 = (x-1)(x+3)(x-2)$$. - Denominator of first fraction: $$x^2 + 4x - 5$$. Factor: $$x^2 + 4x - 5 = (x+5)(x-1)$$. - Numerator of second fraction: $$x^2 + 9x + 20$$. Factor: $$x^2 + 9x + 20 = (x+5)(x+4)$$. - Denominator of second fraction: $$x^3 - 4x^2 + 5x - 2$$. Try to find roots: Test $x=1$: $1 - 4 + 5 - 2 = 0$, so $(x-1)$ is a factor. Divide by $(x-1)$: $$\frac{x^3 - 4x^2 + 5x - 2}{x-1} = x^2 - 3x + 2$$. Factor $x^2 - 3x + 2$: $$x^2 - 3x + 2 = (x-1)(x-2)$$. So, $$x^3 - 4x^2 + 5x - 2 = (x-1)^2 (x-2)$$. 3. **Rewrite the expression with factors:** $$\frac{(x-1)(x+3)(x-2)}{(x+5)(x-1)} \times \frac{(x+5)(x+4)}{(x-1)^2 (x-2)}$$ 4. **Multiply the fractions:** $$\frac{(x-1)(x+3)(x-2)(x+5)(x+4)}{(x+5)(x-1)(x-1)^2 (x-2)}$$ 5. **Cancel common factors:** - Cancel $(x-1)$ in numerator and denominator: $$\frac{\cancel{(x-1)}(x+3)(x-2)(x+5)(x+4)}{(x+5)\cancel{(x-1)}(x-1)^2 (x-2)}$$ - Cancel $(x+5)$: $$\frac{(x+3)(x-2)\cancel{(x+5)}(x+4)}{\cancel{(x+5)}(x-1)^2 (x-2)}$$ - Cancel $(x-2)$: $$\frac{(x+3)\cancel{(x-2)}(x+4)}{(x-1)^2 \cancel{(x-2)}}$$ 6. **Final simplified expression:** $$\frac{(x+3)(x+4)}{(x-1)^2}$$ **Answer:** $$\boxed{\frac{(x+3)(x+4)}{(x-1)^2}}$$