# Real roots of $p(x)=x^n+ax+b$

What can we say about the real roots of $p(x)$?

My Work:

If $n$ is odd I found that $p$ has at most $3$ real roots if $a<0$ and $p$ has at most $1$ real root if $a\geq 0$. How can I classify the roots of $p$ like this when $n$ is even? I got that when $n$ is even $p'(x)=nx^{n-1}+a$. Since $n-1$ is odd $p'(x)$ has exactly one real root. Then $p$ has either $1$ root or $2$ roots or no roots. But how can I classify it exactly? I am stuck. Can anyone please help me?

Note that $p''(x) > 0$, so the $x$ for which $p'(x) = 0$ is a global minimum. Plug in that value of $x$, call it $x_0$, into $p(x)$. If $p(x_0) < 0$, there will be two roots because the minimum goes below the $x$ axis. If $p(x_0) = 0$ there's one root, and if $p(x_0) > 0$ there are none.
• $p''(x) = n(n - 1)x^{n - 2} = 0 \Leftrightarrow x = 0$; otherwise, $p''(x) > 0$. Commented Dec 11, 2014 at 5:54
• If $x_0 = 0$ then $p(x) = x^n + b$, in which case one can directly verify there are two roots if $b < 0$, one if $b = 0$, and none if $b > 0$. This agrees with the above. Commented Dec 12, 2014 at 5:03