Suppose A polynomial $P(x)$, such that,
$$P(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)\cdots$$ having infinite roots. Where $x_1$, $x_2$, $x_3$, $x_4$, $\ldots$ are roots of the polynomial.
My question is what would be the graph of this function? Will it be just the $x$-axis or there will some other curve?
When you have something like this you have to deal with infinite series and radius of convergence and the result is very sensitive to the way in which you define your objects analytically.
I think a good example for this comes from number theory and the theory of partitions. In there the following equation (due to Euler) is proven as a formal expansion:
$\prod_{k = 1}^{\infty}(1 - x^k) = 1 + \displaystyle\sum_{j = 1}^{\infty}(-1)^j(x^{j(3j-1)/2} + x^{j(3j+1)/2})$.
This is a formality between series, without reference to actual values of x, and is obtained byt multiplying the product and collecting powers of $x$ (the proof of it is a different story, and it's called pentagonal number theorem).
Nevertheless you can study both sides and study radius of convergence. For example, with the right hand side, if $|x| \le r < 1$ then we have
$\displaystyle\sum_{j = 1}^{\infty}\left|(-1)^j(x^{j(3j-1)/2} + x^{j(3j+1)/2})\right| \le 2\displaystyle\sum_{j = 1}^{\infty}r^{j(3j-1)/2} \le 2(1 + r + r^2 + ...) = \dfrac{2}{1 - r}$,
which proves this functions converges absolutely and uniformly in compacts of $(-1, 1)$. This means there exists a continuous function to which this series converges and so it makes sense to graph it, but it might not be the graph of a polynomial and it only makes sense in the region of convergence.
Analogously you can check what happens with the product and obtain a similar convergence phenomenon (by expanding the products into polynomials you can basically repeat the argument above). You can see the first products here:
If instead you do it with the polynomials you get (I did it for k = 4):
These are starting to look similar and it indeed seems some function is at the limit indeed.
Notice how all these graphs go through (0, 1) since in both the product and the series version, in the terms that form the sequence these graphs all go thorugh there. On the other hand, in the product version all go through $(\pm, 0)$ since these are zeros of all finite products, meanwhile in the series version it doesnt have those zeroes yet, it just approaches to it in the limit.
Finally, notice how outside $[-1, 1]$ the graphs do not seem to go nowhere, and that makes sense since in there the behaviour at $\pm\infty$ dominates and the degree of the polynomial changes the behaviour.