Algebraists employ formal (vs. functional) polynomials because this yields the greatest generality. Once one proves an identity in a polynomial ring $\rm\ R[x,y,z]\ $ then it will remain true for all specializations of $\rm\:x,y,z\:$ in any ring where the coefficients can be interpreted commutatively, i.e. any ring containing a central image of $\rm\:R,\:$ i.e. any $\rm\:R$-algebra. Thus we can prove once-and-for-all important polynomial identities such as the Binomial Theorem, Cramer's rule, Vieta's formula, etc. and later specialize the indeterminates as need be for applications in specific rings. This allows us to interpret such polynomial identities in the most universal ring-theoretic manner - in the greatest ring-theoretic generality.
For example, when we are solving recurrences over a finite field $\rm\,\mathbb F = \mathbb F_p\,$ it is helpful to employ "operator algebra", working with characteristic polynomials over $\rm\,\mathbb F,\,$ i.e. elements of the ring $\rm\,\mathbb F_p[S]\,$ for $\rm\,S\,$ is the shift operator $\rm\ S\ f(n)\, =\, f(n+1).\,$ They are not polynomial functions on $\rm\,\mathbb F_p,\,$ e.g. generally $\rm\ S^p \ne S\ $ since generally $\rm\ f(n+p) \ne f(n+1).\,$ But any polynomial identity of $\rm\,\mathbb F[x]\,$ specializes to this operator algebra by way of the evaluation map $\rm\,x\mapsto \,S.\,$ E.g. we might specialize universal polynomial factorization identities in order to factor the characteristic polynomial, e.g. difference of squares $\rm\ x^2\! - y^2 = (x\!-\!y)\ (x\!+\!y)\,$ $\,\Rightarrow\,$ $\rm\,S^2\!-\! c^2 = (S\!-\!c)\ (S\!+\! c)\ $ via $\rm\,x,y\mapsto S,c\,,\,$ or perhaps we may specialize a cyclotomic polynomial factorization, etc. This would not be possible if we instead employed the the much less general ring of polynomial functions over $\rm\,\mathbb F\,,\,$ since its specializations of $\rm\,x\,$ must satisfy $\rm\, x^p = x.\,$