Max Real numbers in polynomial

Question

Let $n$ be an odd natural number and

$$P(x) = x^n + ax^2 + b.$$

Is there an option that $P$ would have more than $3$ real roots?

I thought that max of solution will be $n$.

Answer 1

No. Its derivative is $P'(x) = nx^{n-1} + 2ax = x(nx^{n-2} + 2a)$, which has two zeroes, namely $x = 0$ and $x = \sqrt[n-2]{2a/n}$. Therefore $P$ can have at most three zeroes.

As for "max of solution will be $n$", it is the case that the maximal number of solutions for a general polynomial equation $$ x^n + ax^{n-1} + \cdots + bx^2 + cx + d = 0 $$ is $n$, but in this case, you've narrowed it down to a small subset of all such equations (every coefficient except $b$ and $d$ above is set to $0$), and thus there is nothing wrong in the maximal number of solutions beong smaller than $n$.