My book gives this question and in the solution does it following:
Let $y = x^5-5x$. Then $$\frac{dy}{dx} = 5(x-1)(x+1)(x^2+1).$$ By finding the signs of $\frac{dy}{dx}$ in various intervals, it concludes that $x = -1$ is local maximum and $x = 1$ is local minimum. At these points $y$ is equal to $4$ and $-4$ respectively.
Now it says that since $y = -a$, $f(x)$ has three real roots as per the graph, when $a$ is in the range $(-4,4)$. However when $a$ is less than $-4$ or greater than $4$, $f(x)$ has one real root.
Why is this so? If $y = -a$ or $a = -y$, and $y$ lies in the range $(-4,4)$, how can a cross this range? And what does this diagram indicate, and where are the roots in the diagram?
The diagram looks like a graph of $x^5 - 5x$. So, the case when $a = 0$. It is slightly unusual that the $y$ axis is not drawn and there are no grid lines but it still helps.
It shows you that for most of the time, the function is increasing but it has a "wobble" around $0$. It also illustrates the local maximum and minimum that you found.
Changing the value of $a$ will add or subtract uniformly from the value of $y$ and hence it will just move the graph up or down. From the picture, it should be clear that if $a = 4$ (graph moved up by $4$) then the local minimum at $x = 1$ will just touch the the $x$ axis. If $a > 4$ then it will no longer touch and you will have lost a root. Similarly, if $a = -4$ (graph moved down by $4$) then the local maximum at $x = -1$ will now just touch the $x$ axis. If $a < -4$ then you will lose that root.
Redraw the graph with several values of $a$ e.g. $-6, -4, -2, 0, 2, 4, 6$ and see what happens to the roots. You will always have one but sometimes you will have two or three.
This behaviour is very typical of cubic polynomials (maximum power power of $x$ is $3$). Quintic (power $5$) polynomials can be more complicated (up to $5$ roots) but this set is simpler than the general quintic.