Answer
Note that $p^{\prime}(x) \neq 0.$
Assume that p(x) has two zeros and arrive at a contradiction by using Rolle's theorem.
(Please see step-by-step for details)
Work Step by Step
Note that
$p^{\prime}(x)=(2n+1)x^{2n}+a$,
$p^{\prime}(x) > 0$
because the power of x is even, so the term $(2n+1)x^{2n}$ is nonnegative, and $ a > 0 $ .
Let us ASSUME that $p(x)=x^{2n+1}+ax+b$
has more than one real root.
Let $x_{1}$ and $x_{2}$ be two of these.
Since $p(x_{1})=p(x_{2})=0$,
and polynomials are continuous and differentiable everywhere,
by Rolle's Theorem,
there exists $c$ in $(x_{1}, x_{2})$ such that $p^{\prime}(c)=0$.
But $p^{\prime}(x)\neq 0$, so we have a contradiction.
Such a c can not exist.
Our ASSUMPTION was wrong.
Therefore, $p(x)$ can not have two real roots.
