TITLE: Prove that the equation $x^{15} + 7x^{3} - 5 = 0$ has exactly one real solution
QUESTION [5 upvotes]: Prove that the equation $x^{15} + 7x^{3} - 5 = 0$ has exactly one real solution.

A hint that has been given by the teacher is to analyze the function $f(x) = x^{15} + 7x^{3} - 5$.

REPLY [3 votes]: We will use Descartes's rule of signs to determine the number of positive and negative roots of $x^{15}+7x^3-5$. (Zero is obviously not a root.)

The non-zero coefficients of the polynomial per se change sign exactly once. Therefore, it has exactly one positive root.
Replacing $x$ with $-x$ we get $-x^{15}-7x^3-5$, which has no sign changes at all. Therefore the original polynomial has no negative roots.

In conclusion, the original polynomial has exactly one real root, and it is positive.