1. **State the problem:** Simplify the expression $$\frac{10c + c^2 - 3c^3}{5c^2 - 6c - 8}$$. 2. **Rewrite the expression:** Arrange terms in descending powers of $c$ for clarity: $$\frac{-3c^3 + c^2 + 10c}{5c^2 - 6c - 8}$$ 3. **Factor the numerator:** Factor out $c$ first: $$c(-3c^2 + c + 10)$$ Now factor the quadratic inside parentheses: Find two numbers that multiply to $-3 \times 10 = -30$ and add to $1$. These are $6$ and $-5$. Rewrite: $$-3c^2 + 6c - 5c + 10 = (-3c^2 + 6c) + (-5c + 10)$$ Factor each group: $$3c(-c + 2) - 5(c - 2)$$ Rewrite to match factors: $$-3c(c - 2) - 5(c - 2)$$ Factor out $(c - 2)$: $$(c - 2)(-3c - 5)$$ So numerator is: $$c(c - 2)(-3c - 5)$$ 4. **Factor the denominator:** Factor $5c^2 - 6c - 8$. Find two numbers that multiply to $5 \times (-8) = -40$ and add to $-6$. These are $4$ and $-10$ (no, sum is -6, 4 + (-10) = -6). Rewrite: $$5c^2 + 4c - 10c - 8 = (5c^2 + 4c) - (10c + 8)$$ Factor each group: $$c(5c + 4) - 2(5c + 4)$$ Factor out $(5c + 4)$: $$(5c + 4)(c - 2)$$ 5. **Rewrite the entire fraction:** $$\frac{c(c - 2)(-3c - 5)}{(5c + 4)(c - 2)}$$ 6. **Cancel common factors:** $(c - 2)$ appears in numerator and denominator: $$\frac{c\cancel{(c - 2)}(-3c - 5)}{(5c + 4)\cancel{(c - 2)}}$$ 7. **Final simplified expression:** $$\frac{c(-3c - 5)}{5c + 4} = \frac{-3c^2 - 5c}{5c + 4}$$ **Answer:** $$\frac{-3c^2 - 5c}{5c + 4}$$