Respuesta :

1. Factor: 2(2k+1)(k-2)/(k-2)(k+2)(2k+1)
2. Cancel out like terms: 2/(k+2)
3. Final answer: 2/(k+2)

Answer: [tex]\frac{2}{k+2}[/tex]

Explanation:

Firstly, we will split the denominator [tex]k^{2} -4 =(k-2)(k+2)[/tex] by the formula of ([tex]a^{2} -b^{2}[/tex]) = (a-b)(a+b)

then, the given terms [tex]\frac{4k+2}{k^{2}-4} * \frac{k-2}{2k+1}[/tex]

will become [tex]\frac{4k+2}{(k-2)(k+2)} *\frac{k-2}{2k+1}[/tex]

now, cancel the common factor that is k-2 we get

[tex]\frac{4k+2}{(k+2)(2k+1)}[/tex]

Now taking 2 as common from 4k+2 we will get 2(2k+1)

we will get [tex]\frac{2(2k+1)}{(k+2)(2k+1)}[/tex]  common factor will get cancelled which is 2k+1

Hence, we will finally get  [tex]\frac{2}{k+2}[/tex]

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