Solve $3x^5-6x^3=0$ (Presumably using factoring?) I am to solve $3x^5-6x^3=0$
The solutions are provided as 0 and $\pm\sqrt{2}$ and I cannot see how to arrive at this. I wnet down various paths trying to factor the starting equation but that never got me anywhere e.g.
$3x^5-6x^3=0$
$3x^2(x^3-2x)=0$ # now what?
Also tried:
$3x^5-6x^3=0$
$3x(x^4-2x^2)$ # where do I go from here?
Is factoring the right path? How do I know how to approach this problem? How can I solve this equation?

Well, $3x^5-6x^3= 3x^3(x^2-2)$. This gives the real-valued solutions $x_{1,2,3}=0$ and $x_4=\sqrt 2$, $x_5=-\sqrt 2$.
