1. **State the problem:** Simplify the expression $$\frac{x^5-25x^3}{x^5-3x^4-10x^3} - \frac{\frac{x^3-5x^2-6x}{x^2-4x-12}}{\frac{x^2-2x-8}{x^2-x-12}}$$ 2. **Factor all polynomials:** - Numerator of first fraction: $$x^5 - 25x^3 = x^3(x^2 - 25) = x^3(x-5)(x+5)$$ - Denominator of first fraction: $$x^5 - 3x^4 - 10x^3 = x^3(x^2 - 3x - 10) = x^3(x-5)(x+2)$$ - Numerator of second fraction numerator: $$x^3 - 5x^2 - 6x = x(x^2 - 5x - 6) = x(x-6)(x+1)$$ - Denominator of second fraction numerator: $$x^2 - 4x - 12 = (x-6)(x+2)$$ - Numerator of second fraction denominator: $$x^2 - 2x - 8 = (x-4)(x+2)$$ - Denominator of second fraction denominator: $$x^2 - x - 12 = (x-4)(x+3)$$ 3. **Rewrite the expression with factored forms:** $$\frac{x^3(x-5)(x+5)}{x^3(x-5)(x+2)} - \frac{\frac{x(x-6)(x+1)}{(x-6)(x+2)}}{\frac{(x-4)(x+2)}{(x-4)(x+3)}}$$ 4. **Simplify the first fraction by canceling common factors:** $$\frac{\cancel{x^3}\cancel{(x-5)}(x+5)}{\cancel{x^3}\cancel{(x-5)}(x+2)} = \frac{x+5}{x+2}$$ 5. **Simplify the complex fraction:** The complex fraction is: $$\frac{\frac{x(x-6)(x+1)}{(x-6)(x+2)}}{\frac{(x-4)(x+2)}{(x-4)(x+3)}} = \frac{x(x-6)(x+1)}{(x-6)(x+2)} \times \frac{(x-4)(x+3)}{(x-4)(x+2)}$$ Cancel common factors: - In numerator and denominator of first fraction: cancel $(x-6)$ - In numerator and denominator of second fraction: cancel $(x-4)$ - Also note $(x+2)$ appears in denominator of first fraction and denominator of second fraction, so it remains in denominator after multiplication. After cancellation: $$\frac{x\cancel{(x-6)}(x+1)}{\cancel{(x-6)}(x+2)} \times \frac{\cancel{(x-4)}(x+3)}{\cancel{(x-4)}(x+2)} = \frac{x(x+1)}{x+2} \times \frac{x+3}{x+2} = \frac{x(x+1)(x+3)}{(x+2)^2}$$ 6. **Rewrite the original expression with simplified parts:** $$\frac{x+5}{x+2} - \frac{x(x+1)(x+3)}{(x+2)^2}$$ 7. **Find common denominator and subtract:** Common denominator is $(x+2)^2$. Rewrite first term: $$\frac{x+5}{x+2} = \frac{(x+5)(x+2)}{(x+2)^2}$$ So, $$\frac{(x+5)(x+2)}{(x+2)^2} - \frac{x(x+1)(x+3)}{(x+2)^2} = \frac{(x+5)(x+2) - x(x+1)(x+3)}{(x+2)^2}$$ 8. **Expand the numerators:** - Expand $(x+5)(x+2)$: $$x^2 + 2x + 5x + 10 = x^2 + 7x + 10$$ - Expand $x(x+1)(x+3)$: First, $(x+1)(x+3) = x^2 + 3x + x + 3 = x^2 + 4x + 3$ Then multiply by $x$: $$x(x^2 + 4x + 3) = x^3 + 4x^2 + 3x$$ 9. **Substitute back and simplify numerator:** $$x^2 + 7x + 10 - (x^3 + 4x^2 + 3x) = -x^3 + x^2 + 4x + 10$$ 10. **Final simplified expression:** $$\frac{-x^3 + x^2 + 4x + 10}{(x+2)^2}$$ Or equivalently, $$\frac{-(x^3 - x^2 - 4x - 10)}{(x+2)^2}$$ This is the simplified form of the original expression.