# How to factorize this equation

$$\frac{(x^2+1)^2(-2x)-(-x^2+1)(4x^3+4x)}{(x^2+1)^4} = 0$$

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

$$\frac{2x^5-4x^3-6x}{(x^2+1)^4}$$ $$2x^5-4x^3-6x = x^8+4x^6+10x^2+1$$

• First, what is $4x^3+4x$ factorised? Jun 6, 2023 at 20:55
• $-\sqrt3$ should also be a solution, so the solutions are $\pm\sqrt3$ and $0$. Jun 6, 2023 at 20:57
• @SirMrpirateroberts, you should show what you tried in your question, otherwise your question will be closed very soon and the account of the user who has answered your question will be suspended. Jun 6, 2023 at 21:00
• @Angelo Suspension seems rather harsh. Jun 6, 2023 at 21:06
• @Angelo I think you're being a bit overbearing. I'm not sure I'd call this a truly low-quality question. That said: OP: did you intend the RHS to be $0$? If so, you can ignore the denominator completely ($\forall x \in \mathbb{R}: x^2+1 \ne 0$), making this as simple as factoring $x(2x^4-4x^2-6)$, which is quadratic in $x^2$ and factors easily. Can you take it from there? Jun 7, 2023 at 21:30

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?
• @SirMrpirateroberts $(2x^3-6x)=2x(x^2-3)=2x(x-\sqrt{3})(x+\sqrt{3})$ Jun 6, 2023 at 21:17