If you have a polynomial (say $p(x)$) which is irreducible in $\mathbb R[x]$, then is there a practical use of constructing an extension field $\mathbb R[x]/\langle p(x) \rangle$ with a zero rather than just finding the root in $\mathbb C$?
Yes, it's smaller but just because $\mathbb C$ is bigger doesn't mean you have to care about everything there when finding your root.
Is there any practical application for it? Or is it just pure mathematics without any application as yet?
On
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.
You are actually using the theorem that an integral domain quotiented by a maximal ideal is a field. See my answer here .
The main problem is that you have not defined what an algebraic closure is using just the axioms of field theory. Okay sure, for $\Bbb{R}$ you do have $\Bbb{C}$.
But what about arbitrary fields. What about a finite field. How do you get a field where an irreducible polynomial has a root. Say you have an irreducible polynomial like $X^{p}-X+1\in \Bbb{Z}_{p}[X]$. How do you even find a root and attach it to get a "larger" field where this polynomial has a root?
Or even worse, what about function fields? . Say you have $F(X^{n})$ as a field( the field of fractions of the ring $F[X^{n}]$ and you are given an irreducible polynomial like $X^{n}-t^{n}\in F(X^{n})[t]$ . how would you construct a "larger" field where this polynomial has a root?.
That is where this definition comes in handy. You can see my answer in the link . There I have provided a short outline to the process of "attaching a root of a polynomial to a field to get a larger field". All of this is contained in any algebra texts ( I would recomment Dummit and Foote for a reference ).
Edit:- As the conversation in the comment of this answer indicates, I think the main problem is that the op is questioning the need for arbitrary fields in general. So as far as Number theory goes or the study of primes(I don;t think I need to justify it's practical usage in the times of cryptocurrency and stuff like that) , the group $\Bbb{Z}_{p}$ is very very imporant and it is a Field( Called a finite field) . Now as it turns out, every field of characteristic $p$ contains $\Bbb{Z}_{p}$ inside it( to be more precise, an isomorphic copy of it). So this enough should tell you why we should study finite fields . You might learn this as you go forward that all finite fields are just simple extensions of $\Bbb{Z}_{p}$. That is , they occur as $\frac{\Bbb{Z_p}[x]}{(f(x))}$ for some irreducible polynomial. So this should perhaps satisfy your query as to importance of proceeding by quotienting rather than directly moving to $\Bbb{C}$ which as I pointed out is a field of characteristic $0$ and hence does not contain $\Bbb{Z}_{p}$ . See this and this for proofs. This is because we actually do not know explicitly how the root actually looks like. Sure in $\Bbb{R}$ you can say $\sqrt{2}$. But what about the root of the polynomial $x^{2}-2$ over $\Bbb{Z}_{p}[x]$? $\sqrt{2}$ has no meaning here as any extension which will contain this root does not lie in $\Bbb{R}$ as I said that $char\,0 $ fields don't contain $\Bbb{Q}$ and hence no $\Bbb{R}$. The proper mathematical justifications of $\sqrt{2}$ and irrationals are done by Dedekind cuts and the Supremum property but those things are part of Real Analysis .