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?
 A: Well, $3x^5-6x^3= 3x^3(x^2-2)$. This gives the real-valued solutions $x_{1,2,3}=0$ and $x_4=\sqrt 2$, $x_5=-\sqrt 2$.
A: Others have answerd regarding on "how" to solve this problem. Let me take a shot on explaining "why".
At a precalculus level, you have the following tools available:


*

*Factoring polynomials, and

*Solving second grade equations.


These tools are all you need to solve this!
As others (and yourself) have worked out: $$3x^5-6x^3=0 \Rightarrow 3x^3\cdot(x^2-2)=0$$ Now the question is, how to get to the solution, and why it works?
Important: Before revealing the answers below, make sure to work on the reasons yourself.
First of all, ask yourself this: If you have an expression in the form $A \cdot B = 0$ where $A,B\in\mathbb{R}$, what are the values of $A$ or $B$ that let this equality hold? Think on this for a minute before revealing the answer.

 If you have an expression in the form $A \cdot B=0$ where $A,B\in\mathbb{R}$, the equality holds if $A=0$ or $B=0$. 

Now that you know the fact above, you can continue.

 Regarding your problem, you can apply this: $$3x^3\cdot(x^2-2)=0~~~ \Leftrightarrow~~~ 3x^3=0~~\text{or}~~(x^2-2)=0$$

By now, you can solve the full problem. Below is the rest of the answer:

 For the first part the solution is simple: $$3x^3=0\Leftrightarrow x^3=0 \Leftrightarrow x=0$$ So there you have the first value for your solution.  Now, for the second part, $(x^2-2)=0$, you can solve this using any technique you know for solving second grade equations. The easiest way is to factorize this: $$(x^2-2)=0 \Leftrightarrow (x+\sqrt{2})\cdot(x-\sqrt{2})=0$$ Again, this equality holds if any of the two factors are zero: $$(x+\sqrt{2})\cdot(x-\sqrt{2})=0~~~\Leftrightarrow~~~(x+\sqrt{2})=0~~\text{or}~~(x-\sqrt{2})=0$$ So the two possible solutions to this is $x=\sqrt{2}$ and $x=-\sqrt{2}$, or, more compact: $x=\pm\sqrt{2}$.So there you have your solutions: $$3x^5-6x^3=0~~~\Leftrightarrow~~~x=0~~\text{or}~~x=\pm\sqrt{2}$$


All other answers have provided you the solution, but that's only half the issue. If you understand why this solutions work, you will have the capabilities to solve lots of other problems. Always ask yourself: What do I know that could help me solve this problem?, and then try to split the problem into more simple things. That way, you will end up with:


*

*Problems you already know how to solve, and

*Problems that can be split again into more simple ones.


If you iterate enough times using this approach, you'll find your solutions 90% of times... and for the 10% remaining, you can ask for help, knowing that you have done all in your power to solve the original problem.
A: You have $3x^5-6x^3=3x^3(x^2-2)$. Can you take it from here?
A: $$
3x^5−6x^3=0
$$
If you factor out everything you can, then it becomes very easy, since the second factor is a difference of squares:

 \begin{align}3x^3(x^2−2)=0\\3x^3(x-\sqrt{2})(x+\sqrt{2})=0\end{align}

The roots are

 $$0,\quad\pm\sqrt{2},$$

the former with multiplicity $3$, the latter two with multiplicity $1$ each.
