First of all, the expression $\binom{ax+b}{cx+d}$ is never a polynomial when $c\neq 0$. This follows since $\binom{ax+b}{cx+d}\ge \big(\frac{ax+b}{cx+d}\big)^{cx+d}$, and the latter grows exponentially as $x\to\infty$ if $c\neq 0$.
Therefore, you can restrict your attention to expressions like $\binom{ax+b}{k}$. When you expand this out in powers of $x$, the highest degree term is $\frac{a^k}{k!}x^k$, and the constant term is $\binom{b}k$. Therefore, you can determine $a,b$ and $k$ by looking at the leading and constant terms of $f(x)$, plug those into $\binom{ax+b}k$, then see if the result is equal to $f(x)$. If it is, you are done, if not, then $f$ cannot be written in this form.
Not sure What exactly do you mean by factorization, If you want to know a method to convert R.H.S form into L.H.S form, Here is one of the methods
Let $$S= 1 + 2 x + 4 x^2 + 8 x^3 + 16 x^4 + \dots$$
$$Sx= x + 2 x^2 + 4 x^3 + 8 x^4 + 16 x^5 + \dots$$
still the terms don't cancel out, so multiply $Sx$ by 2
$$2Sx= 2 x + 4 x^2 + 8 x^3 + 16 x^4 + \dots$$
now
$$S= 1 + 2 x + 4 x^2 + 8 x^3 + 16 x^4 + \dots (1)$$
$$2Sx= 2 x + 4 x^2 + 8 x^3 + 16 x^4 + \dots (2)$$
subtract (2) from (1)
$$(S-2Sx)=1$$
$$S(1-2x)=1$$
Hence
$$S=\frac{1}{1-2x}$$
which is
$$S=(1-2x)^{-1}$$
Clarification to OP's Further doubts
Let's define $\binom{n}{r}$ as $\binom{n}{r}=\frac{n(n-1)(n-2)....(n-r+1)}{r(r-1)(r-2)....(3)(2)(1)}$ for simplicity
Consider general Binomial expansion
$(1+kx)^n$, Where $k,x,n$ are real number
$(1+kx)^n=\binom{n}{0}(kx)^0+\binom{n}{1}(kx)^1+\binom{n}{2}(kx)^2+\binom{n}{3}(kx)^3+\binom{n}{4}(kx)^4.........$
Now according to OP, It is given that
$f(x)=1+ax+bx^2+cx^3......\tag3$ is indeed a binomial expansion, Which means
$f(x)=\binom{n}{0}(kx)^0+\binom{n}{1}(kx)^1+\binom{n}{2}(kx)^2+\binom{n}{3}(kx)^3+\binom{n}{4}(kx)^4........\tag 4$
Now OP wants to go Backwards i. e from f(x) to Binomial expansion, as it is not always so obvious to identify the coefficients if f(x) is practically given as expanded form of binomial expansion
$f(x)=1+ax+bx^2+cx^3......$
Comparing (3) and (4)
$$k=\frac{coefficient(x)}{\binom{n}{1}}=\frac{coefficient(x)}{n}=\frac{a}{n}$$
$$b=coefficient(x^2)=\frac{k^2(n)(n-1)}{2}$$
$$k^2=\frac{2b}{n(n-1)}$$
solving for $n,k$
$$n=\frac{a^2}{a^2-2b}$$
$$k=\frac{a^2-2b}{a}$$
After finding $k,n$,
One can always write $f(x)=(1+kx)^n$
Applying above on
$$f(x)= 1 + 2 x + 4 x^2 + 8 x^3 + 16 x^4 + \dots$$
$$a=2,b=4$$
it follows
$$k=-2,n=-1$$
$$f(x)=(1+kx)^n=(1-2x)^{-1}$$
Clarification to OP's More interesting doubts
You can Always find All terms using differentiation, Everything we did above can also be analysed by differentiation, Only if it is given that the series is indeed a binomial expansion or product of 2 or more binomial expansions
Let's apply this technique to $$f(x)=1+cx+dx^2+ex^3......=(1+Ax^a)\cdot (1+Bx^b)^q$$
If it is given that the series is indeed a product of two or more binomial expansion
We can Write
$$(1+Ax^a)\cdot (1+Bx^b)^q=1+cx+dx^2+ex^3......$$
Differentiate it one by one and always put $x=0$, Here variables are $A,B,p,q,a,b$
Hence you will need to differentiate 6 times and put $x=0$ again and again to obtain 6 equations, These 6 equations ,when solved will give you Your desired reduced form
If you apply this to $$(1+kx)^n=1+2x+4x^2+8x^3........$$
Differentiate once
$$nk(1+kx)^{n-1}=2+8x+24x^2.......$$
Put $x=0$
$$nk=2\tag 1$$
Differentiate again
$$n(n-1)k^2(1+kx)^{n-2}=8+48x+.....$$
Put $x=0$
$$n(n-1)k^2=8 \tag 2$$
Solving equations $(1)$ and $(2)$ gives you $n=-1$ and $k=-2$
Such things can be done for any number of products of binomial expansion, But with every product the equations of variable formed gets more complex, Also it must be given that $f(x)$ is indeed product of two or more binomial expansions
Best Answer
Let $$P(x)=(1+x)^{4n} +(1+x+x^2)^{2n} +(1+x+x^2+x^3+x^4)^n$$ Then it's reciprocal polynomial is $$\hat{P}(x)=x^{4n}P(1/x)=x^{4n}\left(\left(1+\frac{1}{x}\right)^{4n} +\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^{2n} +\left(1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}\right)^n\right)=P(x)$$ Hence $P(x)$ is a palindromic polynomial and hence $a_r=a_{4n-r}$.