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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.

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  • $\begingroup$ What possibilities does Descartes' Rule of Signs give? $\endgroup$ – Travis Willse Nov 1 '17 at 14:15
  • $\begingroup$ Do you know Sturm's algorithm? $\endgroup$ – Bernard Nov 1 '17 at 14:15
  • $\begingroup$ The derivative is $15 (x-2) (x+2) \left(x^2+2\right)$, the sign of which is very easy to determine. Note $f(-\infty)=-\infty$, $f(-2)>0$, $f(2)<0$, $f(\infty)=\infty$. $\endgroup$ – Gabriel Romon Nov 1 '17 at 14:17
  • $\begingroup$ $1.$ Is your question about this particular polynomial? $2.$ Is your question about these types of polynomial?( I.e. of degree 5, even power absent, or any random odd degree polynomial?)$ 3.$ You want to solve by hand or you can also use a computer program? $\endgroup$ – Jaideep Khare Nov 1 '17 at 14:20
  • $\begingroup$ you can check $f'=0$ it has two roots , so you will have 1-maximum (+) , one minimum(-) ...so you will have $\bf{3}$ real roots $\endgroup$ – Khosrotash Nov 1 '17 at 14:23
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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.

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  • $\begingroup$ thanks, this does make sense. Does this apply to any type of polynomial, e.g. of power 7? $\endgroup$ – Ruben Nov 1 '17 at 14:30
  • $\begingroup$ @Ruben It applies to any polynomial, but it is only useful if it happens to be easy to find all the roots of the derivative. Here it is because the derivative happens to factorise nicely. $\endgroup$ – Especially Lime Nov 1 '17 at 14:59
  • $\begingroup$ sorry to bother you again. And what to do when it does not factorise? Are there any other methods available, when solving without calculator?Thanks. $\endgroup$ – Ruben Nov 1 '17 at 15:15
  • $\begingroup$ @Ruben: what's the meaning of "without calculator"? If you carefully plot a few hundred points by hand, you'll likely get an accurate picture of your polynomial. If you mean formulas, the absolute majority of polynomials of degree give and more have no formulas for their roots. $\endgroup$ – Martin Argerami Nov 4 '17 at 17:12
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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.

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