$x(x^2-2)=0$, The answers are $x = 0, \sqrt{2}$, how do I get there? $$x(x^2-2)=0$$
The answers are $x=\sqrt{2}, 0$ how do I get there?

First begin with noting that your polynomial is of degree $3$. This should tell you that there are $3$ (possible complex) roots. Given $x(x^2 -2) = 0$, it follows that at least one of these factors has to be zero. So possible values are $x = 0$ and $x^2 - 2 =0$. That gives you one solution, $x=0$, the other two follow from a special factorization: $a^2 - b^2 = (a-b)(a+b)$. Using this on the factor $x^2-2$, this gives us : $x^2 - 2 =(x - \sqrt{2}) (x + \sqrt{2}) = 0$. It follows that $(x - \sqrt{2}) = 0$ or $(x + \sqrt{2}) = 0$. Combining this with our first solution $x = 0$, we find that all roots to this equation are $x= 0 , \pm \sqrt{2}$
