$$\frac{(x^2+1)^2(-2x)-(-x^2+1)(4x^3+4x)}{(x^2+1)^4} = 0$$
I am having trouble factorizing this equation such that the solutions are $\sqrt3$,0
I try expanding the terms in the numerator and then expanding the denominator but this did not work out
$$\frac{2x^5-4x^3-6x}{(x^2+1)^4}$$ $$2x^5-4x^3-6x = x^8+4x^6+10x^2+1$$
Let's look at the numerator: $$(x^2+1)^2(-2x)-(-x^2+1)(4x(x^2+1))=(x^2+1)(-2x(x^2+1)-(4x(-x^2+1)))=(x^2+1)(2x^3-6x)$$ Can you take it from here?