Finding the number of real roots in an equation Say I have a polynomial like this one
$$3x^5 -10x^3 -120x +30 = 0$$
And I am asked to find the exact number of real roots.
I have tried  to use the Descartes' Rule of Signs, however, it gives the number of possible roots, but not the exact amount.
How can I solve this question and, in general, for all types of polynomials.
I will be grateful for any help.
 A: There must be a turning point between every two roots. The derivative is $15x^4-30x^2-120=15(x^2-4)(x^2+2)$. So there are only two turning points, at $x=\pm2$, and so at most three real roots. If both turning points are the same side of the $x$-axis there will be one root, and if they are on different sides there will be three, so you just need to check the heights of the two turning points.
A: First of all differentiate the function $f(x) = 3x^5 -10x^3 -120x + 30$:
$$f^{'}(x) = 15x^4 -30x^2 -120$$
Consider the equation $f^{'}(x) = 0$: 
$$15x^4 -30x^2 -120 = 0$$
$$x^4 - 2x^2 -8 = 0$$
$$(x-2)(x+2)(x^2 + 2) = 0$$ 
Hence, there are two turning points at $x = 2$ and $x = -2$. 
Since $f^{'}(x) = 15(x+2)(x-2)(x^2 + 1)$, we can see that $f(x)$ is increasing on $x \le -2$ and on $x \ge 2$ and it's decreasing on $ -2 \le x \le 2$. 
Hence, we have a local maximum at $x = -2$ and a local minimum at $x = 2$
Now, let's have a look at $f(-2)$ and $f(2)$: 
$$f(-2) = 254 > 0$$
$$f(2) = -194 <0$$
Overall, the graph of $f(x)$ would have to intercept with x-axis $3$ times judging by the coordinates of its local maximum and minimum. 
Therefore, you have a total of $3$ real roots. 
