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 $.
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!