5
$\begingroup$

If I have $$x^2(x-3)(x+3)=0$$ then the solutions are: $$x_{1,2}=0, x_3=3, x_4=-3 $$ or $$x_{1}=0, x_2=3, x_3=-3?$$ So are there 4 or 3 solutions?

$\endgroup$

1 Answer 1

14
$\begingroup$

There are $3$ distinct solutions but there are $4$ roots. This has to do with the multiplicity of the roots in a polynomial. It is often times more in vogue to say that there are $3$ solutions to this but, technically, it must be said that there are $3$ distinct roots, as the Fundamental Theorem of Algebra says that there are $d$ roots in $\mathbb{C}$ for a polynomial of degree $d$. Here, $d=4$.

$\endgroup$
2
  • $\begingroup$ So 0 is an even zero of this function and -3 and 3 are odd zeros? Is that correct? $\endgroup$
    – L_McClain
    Commented Dec 1, 2013 at 15:53
  • 1
    $\begingroup$ @user111410 Yes, if parity matters for you. $\endgroup$ Commented Dec 1, 2013 at 15:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .