1. **State the problem:** Multiply the rational functions $$\frac{c^2 - 3c}{c^2 - 25} \cdot \frac{c^2 + 4c - 5}{c^2 - 4c + 3}$$ 2. **Factor all polynomials:** - Numerator of first fraction: $c^2 - 3c = c(c - 3)$ - Denominator of first fraction: $c^2 - 25 = (c + 5)(c - 5)$ (difference of squares) - Numerator of second fraction: $c^2 + 4c - 5 = (c + 5)(c - 1)$ - Denominator of second fraction: $c^2 - 4c + 3 = (c - 3)(c - 1)$ 3. **Rewrite the multiplication with factored forms:** $$\frac{c(c - 3)}{(c + 5)(c - 5)} \cdot \frac{(c + 5)(c - 1)}{(c - 3)(c - 1)}$$ 4. **Multiply the numerators and denominators:** $$\frac{c(c - 3)(c + 5)(c - 1)}{(c + 5)(c - 5)(c - 3)(c - 1)}$$ 5. **Cancel common factors in numerator and denominator:** - Cancel $(c + 5)$: $$\frac{c(c - 3)\cancel{(c + 5)}(c - 1)}{\cancel{(c + 5)}(c - 5)(c - 3)(c - 1)}$$ - Cancel $(c - 3)$: $$\frac{c\cancel{(c - 3)}(c - 1)}{(c - 5)\cancel{(c - 3)}(c - 1)}$$ - Cancel $(c - 1)$: $$\frac{c\cancel{(c - 1)}}{(c - 5)\cancel{(c - 1)}}$$ 6. **Simplified result:** $$\frac{c}{c - 5}$$ **Final answer:** $$\boxed{\frac{c}{c - 5}}$$ This means the product of the two rational functions simplifies to $\frac{c}{c - 5}$, assuming all denominators are not zero (i.e., $c \neq \pm 5, 3, 1$).