Question: I'm studying algebra and I'd like to know why can any quadratic equation be reduced to the product of its two factors.

And why $(x-α) (x-β)$ where $α$ and $β$ are its two roots?

And why is the sum of roots i.e. $α+β=-b/a$ and product of roots $\, αβ = c/a$?

Answer: 
If I understand correctly, then the question asks whether it is possible to define a "multiplication" map $m: \Bbb R^+ \times \Bbb R^+\rightarrow \Bbb R^+$ such that, together with the "addition" map $a: \Bbb R^+ \times \Bbb R^+ \rightarrow \Bbb R^+$ sending $(a, b)$ to $a\cdot b$, makes $\Bbb R^+$ a field.

This is of course possible, in view that $\exp$ is an isomorphism of abelian groups from $(\Bbb R, +)$ to $(\Bbb R^+, \cdot)$.

We simply define the field structure on $\Bbb R^+$ by pushing out the field structure on $\Bbb R$ via the $\exp$ map.

More explicitly, we define $m: \Bbb R^+ \times \Bbb R^+ \rightarrow \Bbb R^+$ by sending $(a, b)$ to $\exp(\ln(a)\cdot \ln(b))$, which is equal to both $a^{\ln (b)}$ and $b^{\ln (a)}$.
The power operator only has a right identity, $1$, and no left identity, so no. It is also not associative nor commutative.

**Edit, to elaborate, since a comment asked for it:**

If an identity element exists in an algebraic structure, it is unique (posit $u$ and $u'$ are different identities for the same magma, you can show they are equal).

$\forall a \in \Bbb R, a^1 = a$ but $1^a = 1$. So the (unique) identity for the power operator is $1$, but it only works on one side (the right side). It is thus not "sufficiently" an identity for it to be the identity of a monoid (*a fortiori* group, ring, field) operator.

Additionally, it is not commutative. In general, $a^b \neq b^a$, such as $2^3 = 8 \neq 9 = 3^2$.

Finally, it is not associative. By convention, we choose to read `a^b^c` as `a^(b^c)`, ie $a^{b^c}$ or $a^{(b^c)}$ (operating from right to left), since `(a^b)^c` corresponds to $(a^b)^c$, or `a^(b*c)`.