Solve the equation $x^2+\frac{4x^2}{(x+2)^2}=5$ Solve the equation $$x^2+\dfrac{4x^2}{(x+2)^2}=5.$$
For $x+2\ne0 \Rightarrow x\ne -2$ we have $$x^2(x+2)^2+4x^2=5(x+2)^2\\x^2(x+2)^2+4x^2-5(x+2)^2=0.$$
Can I factor the LHS? Thank you in advance!
 A: Expand it: $$x^4 +4x^3 +4x^2 +4x^2 -5x^2 -20x -20 =0 \\ x^4+4x^3 +3x^2 -20x-20=0 $$ Then notice that $-1$ is a root and factor: $$(x+1)(x^3+3x^2-20) =0 $$ Now $2$ is a root of the cubic: $$(x+1)(x-2)(x^2+5x+10)=0 $$ I’ll let you finish.
A: Observe what happens when we let $x = u - 1$ to get:
$$\left(u-1\right)^2+\frac{4\left(u-1\right)^2}{\left(u+1\right)^2}-5=0$$
$$\left(u^2-1\right)^2+4\left(u-1\right)^2-5(u+1)^2=0$$
$$u^{4}-2u^{2}+1+4u^{2}-8u+4-5u^{2}-10u-5=0$$
$$u^{4}-3u^{2}-18u=0$$
$$u(u^{3}-3u-18)=0$$
Now by the rational root theorem, the candidate roots for the cubic are the divisors of $18$ up to a sign. We observe that negative candidates all fail because the $-3u$ term is not negative enough to yield negative roots. Checking positive candidates, we have a single root $u = 3$, where there is only 1 positive root by Descartes' rule of signs.
Further factorisation gives $u(u - 3)(u^2 + au + 6)=0$ by comparing coefficients, where $-3 + a = 0 \implies a = 3$. However, the quadratic has discriminant $3^2 - 4 \cdot 1 \cdot 6 < 0$, so the only roots are $x = 0 - 1, 3 - 1$ or $-1, 2$.
