what is the product?\n$\frac{4k + 2}{k^{2}-4}cdot\frac{k - 2}{2k + 1}$\n$\frac{4}{2k+1}$\n$\frac{2}{k…

Explanation:

Step1: Factor the denominator

Factor $k^{2}-4$ using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. Here, $a = k$ and $b = 2$, so $k^{2}-4=(k + 2)(k - 2)$. The expression becomes $\frac{4k + 2}{(k + 2)(k - 2)}\cdot\frac{k - 2}{2k+1}$.

Step2: Factor out the greatest common factor

Factor out 2 from $4k + 2$: $4k+2 = 2(2k + 1)$. The expression is now $\frac{2(2k + 1)}{(k + 2)(k - 2)}\cdot\frac{k - 2}{2k+1}$.

Step3: Cancel out common factors

Cancel out the common factors $(2k + 1)$ and $(k - 2)$ in the numerator and denominator. We get $\frac{2}{k + 2}$.

Answer:

$\frac{2}{k + 2}$