Question

$\frac{6}{z^{2}+4z+4} - \frac{2}{z^{2}-4}$

Explanation:

Step1: Factor denominators

$\frac{6}{(z+2)^2} - \frac{2}{(z+2)(z-2)}$

Step2: Find common denominator

Common denominator: $(z+2)^2(z-2)$

Step3: Rewrite fractions

$\frac{6(z-2)}{(z+2)^2(z-2)} - \frac{2(z+2)}{(z+2)^2(z-2)}$

Step4: Combine numerators

$\frac{6(z-2) - 2(z+2)}{(z+2)^2(z-2)}$

Step5: Simplify numerator

$\frac{6z-12-2z-4}{(z+2)^2(z-2)} = \frac{4z-16}{(z+2)^2(z-2)}$

Step6: Factor numerator

$\frac{4(z-4)}{(z+2)^2(z-2)}$

Answer:

$\frac{4(z-4)}{(z+2)^2(z-2)}$