This is a half hour 3Blue1Brown video: https://youtu.be/NaL_Cb42WyY

If you're willing to watch at least the first half of it, I think it'll give you a very intuitive grounding for 1) why integers have unique polynomial representations, and 2) why that's related to (semi)prime factorization.

Essentially they are both integral domains: https://en.wikipedia.org/wiki/Integral_domain

The space of integers and polynomials have a lot of things in common. They have well-defined addition and multiplication operations, they have a zero value and a one value (for integers it is literally 0 and 1, for polynomials it's the constant function returning 0 and 1, respectively) that obey the usual laws (x + 0 = 0, x1 = x, x0=0, etc). They also both have no zero divisors, i.e. if a*b = 0, then either a=0 or b=0. From just those basic notions you can define divisibility and factorization in a way that applies to both polynomials and integers.

In a very deep way, any property or concept you define on the integers can (often, but of course there are exceptions) be extended to apply to polynomials. There might be some category-theoretic way to formally state that relationship, I'm not sure.

The way positional number systems work, each position corresponds to a power of the base — e.g. 415 = 4 * 10^2 + 1 * 10^1 + 5 * 10^0. A slightly more complicated way to say this is 415 = 4 * x^2 + 1 * x^1 + 5, where x = 10. Ignore the x = 10 part for a bit, this is your polynomial. You can factor this polynomial normally without changing the fact that substituting x = 10 yields the original number. Because of this, it follows that the factors of the polynomial must also be divisors of the original number when you substitute the variable for the appropriate value!

Here's my layman's take - basically it's similar to how computers convert binary integer representations to integers.

For instance, a binary number of 0b1111 could be represented as 2^3 + 2^2 + 2^1 + 2^0 = 15, or 8+4+2+1=15. You now have a typical polynomial which can also be factored algebraically and quickly using something like Mathmatica.

Not all of these binary conversion type polynomials can be factored however as they're irreducible. There's a lot of shifting around in big numbers when you compare the normal binary conversion and what is algebraically viable to be factored.

In my conjecture - it's not clear to me the exact mathematical connection between the normal binary conversion and the perfect polynomial factors we're searching for. I believe it still may be possible to find these unknown polynomial factors by migrating some of the polynomial terms through redistribution of higher value terms and/or performing some sort of hard search. So instead of 2^4+2^1, you can rewrite as 2^3+2^3+2^1, for example.

The factors are exponentially smaller and the search space is reasonable considering that it's basically possible to reduce the polynomial to a large matrix.

I gave up on this path a year ago, so I may be a bit off in my recollection. I was using SymPy (a computer algebra system) but the speed was an issue.

Think about units, tens, hundreds, ..., and stare at step 3 in the article.

In their example of 1111, this binary representation means 2^3 + 2^2 + 2^1 + 2^0. So plugging in 2 to the polynomial x^3 + x^2 + x + 1 gives you the original value.

If the polynomial is factored, plugging 2 in still gives you the same value, but now you have at least 2 integers that multiply to give you the original value, so you've found a factor.