How many real solutions are there for the equation $x^{3}-300x=3000$ How many real solutions are there for the equation $$x^{3}-300x=3000$$

Attempt
There is a solution here I found, without using any algebraic factorization: https://youtu.be/y845hT7aYAQ
and this seems to be most efficient.
But how about one that uses algebraic factorization, or without using derivative?
One method maybe:
$$(x+a)(x+b)(x+c)=x^{3}-300x-3000$$
Then solve $a,b,c$. LHS is
$$ x^{3} + (a+b+c)x^{2} + (ab+ac+bc)x +abc$$
So we have
$$a+b+c=0$$
$$ab+ac+bc=-300$$
$$abc=-3000$$
Using 3rd equation on the 2nd:
$$ -3000(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})=-300$$
Or $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}=\frac{1}{10}$$
Then $a=-(b+c)$ so
$$(b+c)bc=3000$$
By symmetry we also have
$$(a+c)ac=3000$$
$$(b+a)ab=3000$$

By Descartes rule of signs, we have one positive root. Zero isn’t a root.  Now suppose there are negative roots, then using $x \mapsto -x$, these are positive roots of $x^3+3000=300x$.
However by AM-GM,
$x^3+2000+1000> 300\sqrt[3]2\,x>300x$, hence there are no negative roots.
