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!

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.
