1. ## is this correct..

Factoring the following;

$x^3-9x=0$

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

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

= $x=3$ $x=-3$

Thnaks

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

3. Originally Posted by Plato
What happened to $x=0$?
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 $\left( {x + 1} \right)x\left( {x - 1} \right)\left( {x - 3} \right)\left( {x - \pi } \right) = 0$ then the solutions would be
$x = - 1\,,\,x = 0\,,\,x = 1\,,\,x = 3\,,\,\,\& \,x = \pi$.

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