1. Solve $3 \cdot (5x - 8)$: Multiply 3 by each term inside the parentheses: $$3 \cdot 5x = 15x$$ $$3 \cdot (-8) = -24$$ Answer: $$15x - 24$$ 2. Solve $8r \cdot (6r - 2)$: Multiply $8r$ by each term: $$8r \cdot 6r = 48r^2$$ $$8r \cdot (-2) = -16r$$ Answer: $$48r^2 - 16r$$ 3. Solve $5e^4 \cdot (3e^2 - 8e)$: Multiply $5e^4$ by each term: $$5e^4 \cdot 3e^2 = 15e^{4+2} = 15e^6$$ $$5e^4 \cdot (-8e) = -40e^{4+1} = -40e^5$$ Answer: $$15e^6 - 40e^5$$ 4. Solve $-6p \cdot (3p^5 - 6p^2 + 9p)$: Multiply $-6p$ by each term: $$-6p \cdot 3p^5 = -18p^{1+5} = -18p^6$$ $$-6p \cdot (-6p^2) = 36p^{1+2} = 36p^3$$ $$-6p \cdot 9p = -54p^{1+1} = -54p^2$$ Answer: $$-18p^6 + 36p^3 - 54p^2$$ 5. Solve $(-5y^3 + 2y - 9) \cdot 6y$: Multiply $6y$ by each term: $$6y \cdot (-5y^3) = -30y^{1+3} = -30y^4$$ $$6y \cdot 2y = 12y^{1+1} = 12y^2$$ $$6y \cdot (-9) = -54y$$ Answer: $$-30y^4 + 12y^2 - 54y$$ 6. Solve $\frac{1}{4} w^2 \cdot \left(\frac{4}{5} w^2 - w\right)$: Multiply $\frac{1}{4} w^2$ by each term: $$\frac{1}{4} w^2 \cdot \frac{4}{5} w^2 = \frac{1}{4} \cdot \frac{4}{5} w^{2+2} = \frac{4}{20} w^4 = \frac{1}{5} w^4$$ $$\frac{1}{4} w^2 \cdot (-w) = -\frac{1}{4} w^{2+1} = -\frac{1}{4} w^3$$ Answer: $$\frac{1}{5} w^4 - \frac{1}{4} w^3$$ 7. Solve $(3m^2 - 6m + 2)(-5m)$: Multiply $-5m$ by each term: $$-5m \cdot 3m^2 = -15m^{1+2} = -15m^3$$ $$-5m \cdot (-6m) = 30m^{1+1} = 30m^2$$ $$-5m \cdot 2 = -10m$$ Answer: $$-15m^3 + 30m^2 - 10m$$ 8. Solve $-3u^3 \cdot (-7u^3 + 3u - 14)$: Multiply $-3u^3$ by each term: $$-3u^3 \cdot (-7u^3) = 21u^{3+3} = 21u^6$$ $$-3u^3 \cdot 3u = -9u^{3+1} = -9u^4$$ $$-3u^3 \cdot (-14) = 42u^3$$ Answer: $$21u^6 - 9u^4 + 42u^3$$ 9. Solve $\frac{3}{4} f^2 \cdot \left(\frac{8}{13} f^2 - \frac{4}{9} f + 3\right)$: Multiply $\frac{3}{4} f^2$ by each term: $$\frac{3}{4} f^2 \cdot \frac{8}{13} f^2 = \frac{3}{4} \cdot \frac{8}{13} f^{2+2} = \frac{24}{52} f^4 = \frac{6}{13} f^4$$ $$\frac{3}{4} f^2 \cdot \left(-\frac{4}{9} f\right) = -\frac{12}{36} f^{2+1} = -\frac{1}{3} f^3$$ $$\frac{3}{4} f^2 \cdot 3 = \frac{9}{4} f^2$$ Answer: $$\frac{6}{13} f^4 - \frac{1}{3} f^3 + \frac{9}{4} f^2$$ 10. Solve $\left(-\frac{3}{5} i^2 - \frac{5}{8} i^2 + 17\right) \cdot \left(-\frac{1}{3} i^4\right)$: First combine like terms inside the parentheses: $$-\frac{3}{5} i^2 - \frac{5}{8} i^2 = \left(-\frac{24}{40} - \frac{25}{40}\right) i^2 = -\frac{49}{40} i^2$$ So expression becomes: $$\left(-\frac{49}{40} i^2 + 17\right) \cdot \left(-\frac{1}{3} i^4\right)$$ Multiply each term: $$-\frac{49}{40} i^2 \cdot -\frac{1}{3} i^4 = \frac{49}{120} i^{2+4} = \frac{49}{120} i^6$$ $$17 \cdot -\frac{1}{3} i^4 = -\frac{17}{3} i^4$$ Answer: $$\frac{49}{120} i^6 - \frac{17}{3} i^4$$