Let $n \in \mathbb N$ and $a,b \in \mathbb R$. Prove or disprove that $x^n+ax+b=0$ has no more than 3 solutions.

Question

Let $n \in \mathbb N$ and $a,b \in \mathbb R$. Prove or disprove that $x^n+ax+b=0$ has no more than 3 solutions.

I believe this statement is true.

Let's set $f(x)=x^n+ax+b$ and since it's a polynomial, it's continuous and differentiable.

$$f'(x)=nx^{n-1}+a$$ $$f''(x)=n(n-1)x^{n-2}$$

My thoughts:

I believe that if the polynomial has n solutions, then the first derivative has at most n-1 solutions, second has at most n-2 and so on. Since our second derivative has 1 solution, and our second derivative has 1 solution as well, then that means we have have at most 3 solutions.

I feel that this train of thought isn't rigorous enough for a proof.

Answer 1

Your logic is nearly correct. If a differentiable function $f$ has $n$ distinct zeros, its derivative $f'$ must have at least (not at most) $n-1$ distinct zeros by Rolle's Theorem. By the same logic, $f''$ must have at least $n-2$ distinct zeros.

If your function $f$ had $4$ (or more) distinct zeros, its second derivative $f''$ would have at least $2$ distinct zeros, which it clearly does not.