Finding the number of (real) solutions of polynomials and their location on the x-axis I've stumbled upon some questions while practicing polynomials. For each question, I can find the solution, but I think there should be an easier way.
For example: $x^7 + x + 1 = 0$. I'm able to find the (real) solutions of this equation by using methods like Horner's. However, is there an easier way to find the number of real solutions this equation has. I don't need to know the real solutions, just the number of them.
Next: $x^4-4x^3+2x^2+2x-1=0$. I have given that their are $4$ real solutions. The question: How many of those solutions are smaller than $0$? Ofcourse I can compute them all, but is their a faster, more intuitive way for finding the answer to this question?
Lastly, how can I determine how many polynomial functions exist through some given points. For example, how many (polynomial) functions do their exist through the points: $(0, 0)$, $(1, 1)$, $(2, 4)$ and $(3, 9)$. My intuition says one solution $y = x^2$, but is their a way to calculate this using arithmetic? Are their overal rules/exercises to improve intuition for this kind of functions and their roots/real solutions? Thank you in advance! PS: I'm not a native English speaker, so some translations/terminology may be wrong, don't hesitate to correct and/or ask what I mean if it's not clear.
 A: *

*I doubt there is general simple way, however for $x^7 + x + 1$ we can notice that it has at least one real root (as polynomial of odd degree), and it's derivative $7x^6 + 1  = 0$ has no real roots, so the polynomial has at most one real root (if a polynomial has two real roots, there is a root of it's derivative between them, and if it has a root of multiplicity greater than $1$, it's also root of it's derivative).

*You can use Descarte's rule of signs. There are $3$ sign changes, and you know all roots are real, so number of positive roots is equal to number of sign changes.

*There are always infinitely many polynomial functions passing through given finite set of points (assuming $x$ coordinates of points are distinct of course). In your example, you can add any polynomial of form $p(x) \cdot x \cdot (x - 1) \cdot (x - 2) \cdot (x - 3)$ to any solution, and get a new solution.
However, if you have $n$ points, there is exactly one polynomial of degree $< n$ passing through them (because difference of two such polynomials will necessary have $n$ roots, and polynomial of degree $n - 1$ with $n$ roots is zero polynomial).
