Why do people study algebraic extension?

I think the original motivation for studying field extensions was, as in the theorem you stated, to solve polynomials. One of the big results after a few lectures of algebraic field extensions is that every field can be embedded into a unique algebraically closed field, called the algebraic closure.

Actually, solving equations is really the motivation for all the historical expansions of the concept "number." Think of it this way: we have the natural numbers, $1,2,3,\ldots$ and we can solve equations like $x+2 = 4$. But then we can pose equations like $x+ 2 = 2$ and $x+4 = 2$, so we want to extend our number system to include solutions to these, so we add zero and negative integers.

Then we notice we can pose equations like $2x = 4$, which has a nice solution $x=2$, and also $4x = 2$, which doesn't have a solution in the integers. So again, we extend our number system again to include things like $\frac 12$. Now we can solve any linear equation $ax+b = 0$ where $a,b \in \mathbb{N}$.

We also have equations like $x^2 - 2 = 0$, and so we start adding irrational numbers like $\sqrt{2}$. We can complete the rational numbers to form the field of real numbers $\mathbb{R}$. But still, we have equations we can't solve, like $x^2 + 1 = 0$. To get a solution of this, we add the number $i = \sqrt{-1}$ and get the complex numbers. Going from $\mathbb{R}$ to $\mathbb{C}$ this way is an algebraic field extension.

As the theorem you stated says, this procedure is much more general than just $\mathbb{R}$ to $\mathbb{C}$: it says that given any abstract field $F$ and any polynomial equation with coefficients from that field, we can enlarge the "number system" that is $F$ to include a solution.

EDIT: As was pointed out in a comment, the algebraic closure of a field is only unique up to a non-canonical isomorphism.

It's important in algebraic number theory. Without going into too much detail, many results about the ring $\mathbb{Z}$ of integers are easier to obtain and understand when we generalize the notion of an integer. This is done by replacing the field $\mathbb{Q}$ of rational numbers with a field $K$ of the form $\mathbb{Q}(a)$, where $a$ is a complex number which is algebraic over $\mathbb{Q}$.

For the field $\mathbb Q$ of rational numbers, the ring of integers is of course $\mathbb{Z}$.

For the field $\mathbb{Q}(i) = \{ a + bi : a, b \in \mathbb{Q}\}$, the ring of integers is $\mathbb{Z}[i] = \{a + bi : a, b \in \mathbb{Z} \}$. This is also called the ring of Gaussian integers.

More generally, every field $K$ containing $\mathbb{Q}$ of the form $\mathbb{Q}(a)$ as above, has its own "ring of integers."

Here is a classical theorem about prime numbers:

Theorem: Let $p \geq 3$ be a prime number. There exist integers $x, y \in \mathbb{Z}$ such that $p = x^2 + y^2$ if and only if $p - 1$ is divisible by $4$.

This theorem was known before people were working with the ring $\mathbb{Z}[i]$, but it is much easier to prove using the ring $\mathbb{Z}[i]$. Basically, one defines the notion of an element of $\mathbb{Z}[i]$ being prime. Then the question becomes to show which prime numbers in $\mathbb{Z}$ remain prime when regarded as an element of $\mathbb{Z}[i]$.

To add to Joshua Ruiter's answer.

You can teach a computer how to calculate with extensions field "exactly". If you see the $\sqrt{2}$ as an infinite decimal, your computer can only compute approximations (in finite time). But if you teach the computer that it is a pair $(0,1)$, with some algebraic properties, it will be able to give back exact answers to your calculations that involve $\sqrt{2}$.