is this correct..

**Morzilla** · Jan 20th 2008, 01:24 PM

Factoring the following;

$x^3-9x=0$

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

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

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

Thnaks

**Plato** · Jan 20th 2008, 01:41 PM

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 $x=0$ ?

**Morzilla** · Jan 20th 2008, 01:48 PM

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?

**Plato** · Jan 20th 2008, 02:05 PM

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
$x = - 1\,,\,x = 0\,,\,x = 1\,,\,x = 3\,,\,\,\& \,x = \pi$ .

**Morzilla** · Jan 20th 2008, 02:40 PM

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!

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