Number of zeroes in a particular interval [-1,1] for $x^{2n+1} + (2n + 1) x + a = 0$ 
Let n be a natural number and let a be a real number. The number of
  zeros of $x^{2n+1} + (2n + 1) x + a = 0$ in the interval $[-1, 1]$ is ?

The options given are:

(A) 2 if a > 0 

(B) 2 if a < 0 

(C) at most one for every vale of a 

(D) at least three for every value of a

Hints please.I can't understand how to solve.
 A: Let
$p(x) = x^{2n + 1} + (2n + 1)x + a; \tag{1}$
then
$p'(x) = (2n + 1) x^{2n} + (2n + 1).  \tag{2}$
We note that for all real $x$ and all integers $n \ge 0$,.
$x^{2n} \ge 0 \tag{3}$
and
$2n + 1 > 0; \tag{4}$
thus we have
$p'(x) > 0 \tag{5}$
for all $x \in \Bbb R$.  We accept as known the fact that $p(x)$, being of odd degree, has at least one zero, $x_0 \in \Bbb R$; in fact, since $p'(x) > 0$ everywhere, it follows that $x_0$ is the only $0$ of $p(x)$; for $x > x_0$,  $p(x) > 0$, and for $x < x_0$, $p(x) < 0$.  This of course follows from the fact that $p'(x) > 0$ implies that  $p(x)$ is a strictly monotonically increasing function $x$; indeed, if $x_2 > x_1$ we have
$p(x_2) - p(x_1) = \int_{x_1}^{x_2} p'(s) ds > 0, \tag{6}$
whence $p(x_2) > p(x_1)$.
In any event, we see that, since $p(x)$ has precisely one zero in all of $\Bbb R$, the only remaining issue is if $x_0 \in [-1, 1]$ or not; thus the correct answer is (C).
A: To get an idea, compare the polynomial to the product of its dominant binomials. For $|a|\ll 2n+1$ this is
$$
(x^{2n}+(2n+1))(x+\tfrac{a}{2n+1})=x^{2n+1}+(2n+1)x+a\ +\ \tfrac{a}{2n+1}x^2n
$$
By perturbation arguments, the first factor gives only non-real roots that stay non-real and have absolute value close to $\sqrt[2n]{2n+1}>1$. The second factor gives exactly one root inside $[-1,1]$.
For $|a|\gg (2n+1)$, the dominating binomial is $$x^{2n+1}+a$$ which has exactly one real root $\sqrt[2n+1]{-a}$ outside $[-1,1]$. This stays stable under the (relatively) small perturbation needed to reach the given polynomial.

To get a definitive result, consider that, as per A.S. and as usually done in such trinomial tasks,
$$
f'(x)=(2n+1)(x^{2n}+1)\ge 2n+1>0,
$$
which tells about monotonicity and thus about the maximum number of roots.
