which expression is equivalent to $\frac{c^{2}-4}{c + 3}div\frac{c + 2}{3(c^{2}-9)}$?\n$\frac{c +…

which expression is equivalent to $\frac{c^{2}-4}{c + 3}div\frac{c + 2}{3(c^{2}-9)}$?\n$\frac{c + 3}{c^{2}-4}div\frac{c + 2}{3(c^{2}-9)}$\n$\frac{c^{2}-4}{c + 3}div\frac{3(c^{2}-9)}{c + 2}$\n$\frac{c^{2}-4}{c + 3}cdot\frac{3(c^{2}-9)}{c + 2}$\n$\frac{c + 3}{c^{2}-4}cdot\frac{c + 2}{3(c^{2}-9)}$
Answer
Explanation:
Step 1: Convert division to multiplication by reciprocal
$\frac{c^2 - 4}{c + 3} \div \frac{c + 2}{3(c^2 - 9)} = \frac{c^2 - 4}{c + 3} \times \frac{3(c^2 - 9)}{c + 2}$
Step 2: Factor quadratic expressions
$c^2 - 4 = (c - 2)(c + 2)$, $c^2 - 9 = (c - 3)(c + 3)$
Substitute: $\frac{(c - 2)(c + 2)}{c + 3} \times \frac{3(c - 3)(c + 3)}{c + 2}$
Step 3: Cancel common factors
Cancel $(c + 2)$ and $(c + 3)$: $\frac{(c - 2)\cancel{(c + 2)}}{\cancel{c + 3}} \times \frac{3(c - 3)\cancel{(c + 3)}}{\cancel{c + 2}} = 3(c - 2)(c - 3)$
Answer:
C. $\frac{c^2 - 4}{c + 3} \cdot \frac{3(c^2 - 9)}{c + 2}$