If you ask about fields $k$ other than $\mathbb{R}$, then I think I can give several good answers to this question.
(1) As other answers have pointed out, you may not have proved that $k$ has an algebraic closure yet.
(2) Even if you know that $k$ has an algebraic closure (perhaps $k = \mathbb{Q}$), computing in $k[x]/f(x)$ may be more convenient than computing in $k^{alg}$. For example, if I had to do a lot of computations in $\mathbb{Q}[\zeta_5]$, where $\zeta_5$ is a primitive $5$-th root of unity, I'd rather store my numbers in the form $a_1 \zeta+a_2 \zeta_2+a_3 \zeta^3+a_4 \zeta^4$ for $a_1$, $a_2$, $a_3$, $a_4 \in \mathbb{Q}$ than storing them as $x+iy$ with $x$ and $y$ floating point. (There are tradeoffs, though: As the degree of $f$ gets larger, floating point gets more useful.)
Point (2) is much more true if $k$ is not a subfield of $\mathbb{C}$. In that case, basically* the only way to compute in $k^{alg}$ is to work in finitely generated subfields of it.
(3) If you are heading for Galois theory, you want to show that, if $f(x)$ is an irreducible polynomial, and $K$ is a splitting field of $f$, then $\text{Aut}(K/k)$ acts transitively on the roots of $f$. To be concrete, you want to show that there is an automorphism of $\mathbb{Q}(\sqrt[3]{2}, \omega)$ taking $\sqrt[3]{2}$ to $\omega \sqrt[3]{2}$. This is pretty straightforward if you write an element of $\mathbb{Q}(\sqrt[3]{2}, \omega)$ as $a + b \omega + c \sqrt[3]{2} + d \omega \sqrt[3]{2} + e \sqrt[3]{4} + f \omega \sqrt[3]{4}$, for $a$, $b$, $c$, $d$, $e$, $f \in \mathbb{Q}$. It is hard (in my opinion) if you think of $\mathbb{Q}(\sqrt[3]{2}, \omega)$ as the subfield of $\mathbb{C}$ generated by $\sqrt[3]{2}$ and $\omega$. In particular, you need the Axiom of Choice if you want to prove this automorphism extends to $\mathbb{C}$.
If you insist on thinking about $\mathbb{R}$, the advantages are much less, because you'll be using floating point numbers to store the elements of $\mathbb{R}$ any way, and because $\mathbb{C}$ is so much more concrete than a general $k^{alg}$.
The one reason I can think of is that perhaps you haven't proved the FTA yet. There are proofs of the FTA where you want to first know that there is some field where $f(x)$ has roots, before proving that $\mathbb{C}$ is such a field. Several mathematicians (Euler, Lagrange, Laplace) gave flawed proofs of the FTA in the 18th century which assumed such a field existed. From a modern perspective, there is a slick Galois theory proof of the FTA; if you want to take this route, you need to be able to build splitting fields without the FTA.
* For experts, the "basically" is my way of indicating that yes, I am aware of Kedlaya and other's work on explicitly working in an algebraic closure of $\mathbb{F}_p[t]$ by generalized Puiseux series.
Also are you sure you want to deal with polynomials in R[x] and not in Q[x]?- my point is why should I care at all where the root exists as long as I can find the root – user93353 Apr 18 '22 at 21:21further algebraically one can't distinguish between the two roots $\pm\sqrt{2}$- in all practical uses of finding the roots of a polynomial like plotting the curve, you do need to distinguish between the two roots. Which goes back to my question as to the practical use of finding the root of the polynomial this way. I am just trying to step back a little & think as to why we are into finding roots at all in the first place? – user93353 Apr 18 '22 at 21:43