1. Planteamos el problema: calcular el producto $A(x) \cdot (-D(x))$ donde $$A(x) = 5x^5 + 3x^4 - 4x^2 + \frac{x}{2} - 2$$ $$D(x) = \frac{x^2}{5} + 2x + 2$$ 2. Primero, calculamos $-D(x)$ cambiando el signo a cada término: $$-D(x) = -\left(\frac{x^2}{5} + 2x + 2\right) = -\frac{x^2}{5} - 2x - 2$$ 3. Ahora multiplicamos $A(x)$ por $-D(x)$ término a término: $$A(x) \cdot (-D(x)) = \left(5x^5 + 3x^4 - 4x^2 + \frac{x}{2} - 2\right) \cdot \left(-\frac{x^2}{5} - 2x - 2\right)$$ 4. Multiplicamos cada término de $A(x)$ por cada término de $-D(x)$ y sumamos: - $5x^5 \cdot \left(-\frac{x^2}{5}\right) = 5x^5 \cdot -\frac{x^2}{5} = -\cancel{5}x^{5+2} / \cancel{5} = -x^7$ - $5x^5 \cdot (-2x) = -10x^{6}$ - $5x^5 \cdot (-2) = -10x^5$ - $3x^4 \cdot \left(-\frac{x^2}{5}\right) = 3x^4 \cdot -\frac{x^2}{5} = -\frac{3}{5}x^{6}$ - $3x^4 \cdot (-2x) = -6x^{5}$ - $3x^4 \cdot (-2) = -6x^{4}$ - $-4x^2 \cdot \left(-\frac{x^2}{5}\right) = -4x^2 \cdot -\frac{x^2}{5} = \frac{4}{5}x^{4}$ - $-4x^2 \cdot (-2x) = 8x^{3}$ - $-4x^2 \cdot (-2) = 8x^{2}$ - $\frac{x}{2} \cdot \left(-\frac{x^2}{5}\right) = \frac{x}{2} \cdot -\frac{x^2}{5} = -\frac{1}{10}x^{3}$ - $\frac{x}{2} \cdot (-2x) = -x^{2}$ - $\frac{x}{2} \cdot (-2) = -x$ - $-2 \cdot \left(-\frac{x^2}{5}\right) = \frac{2}{5}x^{2}$ - $-2 \cdot (-2x) = 4x$ - $-2 \cdot (-2) = 4$ 5. Sumamos todos los términos semejantes: - $x^7$: $-x^7$ - $x^6$: $-10x^6 - \frac{3}{5}x^6 = -\frac{50}{5}x^6 - \frac{3}{5}x^6 = -\frac{53}{5}x^6$ - $x^5$: $-10x^5 - 6x^5 = -16x^5$ - $x^4$: $-6x^4 + \frac{4}{5}x^4 = -\frac{30}{5}x^4 + \frac{4}{5}x^4 = -\frac{26}{5}x^4$ - $x^3$: $8x^3 - \frac{1}{10}x^3 = \frac{80}{10}x^3 - \frac{1}{10}x^3 = \frac{79}{10}x^3$ - $x^2$: $8x^2 - x^2 + \frac{2}{5}x^2 = \left(8 - 1 + \frac{2}{5}\right)x^2 = \left(7 + \frac{2}{5}\right)x^2 = \frac{35}{5}x^2 + \frac{2}{5}x^2 = \frac{37}{5}x^2$ - $x$: $-x + 4x = 3x$ - Constante: $4$ 6. Resultado final: $$A(x) \cdot (-D(x)) = -x^7 - \frac{53}{5}x^6 - 16x^5 - \frac{26}{5}x^4 + \frac{79}{10}x^3 + \frac{37}{5}x^2 + 3x + 4$$