Descartes`s Rule in subject GRE exam.

Question

I am solving this question:

And I found the following answer

I do not understand why there is one sign change in $f$, could anyone explain this for me?

Answer 1

The first two coefficients have a positive sign and the third one has a negative sign. Therefore there is exactly one sign change.

Answer 2

$2x^5+8x-7$ is an increasing function and odd degree polynomials have at least one root so the answer is $1$.

Answer 3

While the book is doing an IMHO terrible job of explaining it, the number of 'sign changes' being referenced is not the sign changes in the function itself, but rather the number of sign changes in the sequence of coefficients $\langle a_n, a_{n-1}, \ldots, a_1, a_0\rangle$ (note that zero coefficients need to be removed here). In the case under offer here, the sequence is is $\langle 2, 8, -7\rangle$; this has one sign change when going from $8$ to $-7$, so the maximum number of positive roots of the function is $1$. Descartes's rule also says that the actual number of roots varies from this value by an even number; in other words, it can be two less than this, four less, etc. But in this case, two less is already $-1$ and it's impossible to have a negative number of roots, so the number of positive roots of the function must be exactly $1$.

To find the maximum number of negative roots, we apply the same rule to the function $f(-x)$; as the book notes, this function has sequence $\langle -2, -8, -7\rangle$ and there are no sign changes in this sequence (all the terms are negative) so there can be no negative zeroes. (This technique tells you nothing about whether $0$ itself is a zero of the polynomial, but that's true if and only if the constant coefficient of the polynomial is zero, which it isn't here).

For more information, I recommend reading either the Wikipedia or Mathworld pages on Descartes' sign rule.