Finding the number of (real) solutions of polynomials and their location on the x-axis

Question

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.

Answer 1

1. 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).

2. 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.

3. 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).