In the article that proves Descartes' Rule of Signs:

https://www.cut-the-knot.org/fta/ROS2.shtml

there are 2 basic statements:

Let’s denote the number of variations in sign of the coefficients of f(x) by V, and denote the number of positive real roots of f(x) by P.

I. Let f(x) = x^n + (an-1)*x^(n-1) + … + a1*x + a0. If a0 < 0, then V is odd; if a0 > 0, then V is even.

II. Let f(x) = x^n + (an-1)*x^(n-1) + … + a1*x + a0. If a0 < 0, then P is odd; if a0 > 0, then P is even.

The provided proof for II is clear enough. However, for I there is a problem: it is evidently true only if coefficients (an-1) to a1
are positive, since the odd/even condition would depend only on a0. Nothing is said to prove that it's true for other cases. For example, if all coefficients are positive, except a1 and a0, (contiguous coefficients),V=odd, statement is true, but if a2 and a0 are negative, statement would fail.

Any suggestion?

Thanks

No, it would not. There will be changes of sign between

1) a₃ and a₂ (+ → -)
2) a₂ and a₁ (- → +) and
3) a₁ and a₀ (+ → -).

Have a look at the following apparently not relevant page:

https://www.cut-the-knot.org/Curriculum/Algebra/FirstProof.shtml

The proof is simple enough.

Take any sequence of signs plus and minus. Count the number of sign changes. Then change one of the signs. If this one is at an end, then obviously the parity of the sign changes will change. But if you change any sign in the middle, it will not. Just by inspection, what matters (if at all) are the signs of the immediate neighbors of the sign that you change. So let's consider eight cases:

+++ (0) → +-+ (2)
++- (1) → +-- (1)
+-+ (2) → +++ (0)
+-- (1) → ++- (1)
-++ (1) → --+ (1)
-+- (2) → --- (0)
--+ (1) → -++ (1)
--- (0) → -+- (2)