1. ## is this correct..

Factoring the following;

$\displaystyle x^3-9x=0$

= $\displaystyle x(x^2-9)=0$

= $\displaystyle (x-3)(x+3)=0$

=$\displaystyle x=3$ $\displaystyle x=-3$

Thnaks

2. Originally Posted by Morzilla
Factoring the following;
$\displaystyle x^3-9x=0$
= $\displaystyle x(x^2-9)=0$
= $\displaystyle (x-3)(x+3)=0$
=$\displaystyle x=3$ $\displaystyle x=-3$
What happened to $\displaystyle x=0$?

3. Originally Posted by Plato
What happened to $\displaystyle x=0$?
so are you saying that the x in x(x-3)(x+3) should also be accounted for?

4. Originally Posted by Morzilla
so are you saying that the x in x(x-3)(x+3) should also be accounted for?
Well of course!
If you had $\displaystyle \left( {x + 1} \right)x\left( {x - 1} \right)\left( {x - 3} \right)\left( {x - \pi } \right) = 0$ then the solutions would be
$\displaystyle x = - 1\,,\,x = 0\,,\,x = 1\,,\,x = 3\,,\,\,\& \,x = \pi$.

5. Originally Posted by Plato
Well of course!
If you had $\displaystyle \left( {x + 1} \right)x\left( {x - 1} \right)\left( {x - 3} \right)\left( {x - \pi } \right) = 0$ then the solutions would be
$\displaystyle x = - 1\,,\,x = 0\,,\,x = 1\,,\,x = 3\,,\,\,\& \,x = \pi$.
......ohhhhhhh! I see, I see ok thank you very much!