Thread: 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

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 $\displaystyle x=0$?

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?

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 $.

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!