# Find zeros of function $f(x)$

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?

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, $d=4$.