what is the completely factored form of $x^{4}y - 4x^{2}y - 5y$?\n$y(x^{2}-5)(x^{2}+1)$\n$y(x^{2}+5)(x^{2}-1)…

Explanation:

Step1: Factor out the common factor

Factor out $y$ from $x^{4}y - 4x^{2}y - 5y$. We get $y(x^{4}-4x^{2}-5)$.

Step2: Let $t = x^{2}$

Let $t=x^{2}$, then the expression inside the parentheses becomes $t^{2}-4t - 5$.

Step3: Factor the quadratic expression

Factor $t^{2}-4t - 5$. We need two numbers that multiply to $- 5$ and add up to $-4$. The numbers are $-5$ and $1$. So $t^{2}-4t - 5=(t - 5)(t + 1)$.

Step4: Substitute back $t=x^{2}$

Substitute $t = x^{2}$ back into $(t - 5)(t + 1)$, we get $(x^{2}-5)(x^{2}+1)$. So the factored - form of $x^{4}y-4x^{2}y - 5y$ is $y(x^{2}-5)(x^{2}+1)$.

Answer:

A. $y(x^{2}-5)(x^{2}+1)$