So you have x^3 - 4x = 0. What you can do is pull out an x from both x^3 and - 4x so it looks like this:
![x( {x}^(2) - 4) = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/49jkhhsoc4onoxh0arj46xcd25d8gsyewe.png)
Then you can find a number that makes the part inside the parentheses turn into zero. For beginners, it may be easier to write it out seperately and solve for x.
![{x}^(2) - 4 = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/l9joogtx46xf9rmvw7gmhcqrjffya7csnp.png)
We need to solve for x, so the first step is to add 4 to both sides, so we get something like this:
![{x}^(2) = 4](https://img.qammunity.org/2020/formulas/mathematics/middle-school/g3dhbzrxrg6fk2tp8yx5x865m8rwst73xx.png)
Then, we can square root both sides to get rid of the power on the x, so it looks like this:
![x = √(4)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ox6jdouvv97t99upxsfn093n46pk5xk2rc.png)
Now, every square root has two answers, a positive and a negative. If we look at the bottom example:
![{2}^(2) = 4](https://img.qammunity.org/2020/formulas/mathematics/middle-school/94q34hryrelxm4zrrevs9auemvxkskzi6b.png)
![{( - 2)}^(2) = 4](https://img.qammunity.org/2020/formulas/mathematics/middle-school/myw2y8qqvlotut2z1ux40wa0kqz72m24u7.png)
We can see that both -2 and 2 to the power of two will equal to 4.
So finally, we get:
![x = - 2 \: and \: 2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/20dfmekj41l236rmd2dslylue1u7sj8ops.png)
These are the other 'Zero's for the original function. If you are not sure of what a 'Zero' is, it is where the function crosses over the x-axis on a graph.