Why do we not consider the case where $u$ and $v$ in the factored form $(x+u)(x+v)$ are fractions? When factoring the quadratic $x^2 + bx + c $, where $b$ and $c$ are integers, why do we not consider the case where $u$ and $v$ in the factored form $(x+u)(x+v)$ are fractions?
We wish to write this as $(x + u) (x + v)$. Where $u = \frac{p}{q}$ and $v =\frac{m}{n}$.
\begin{align}
\left(x + \frac{p}{q}\right)\left(x + \frac{m}{n}\right)&= x\left(x+\frac{m}{n}\right)+\frac{p}{q}\left(x + \frac{m}{n}\right)\\
&=x^2 + \frac{mx}{n} + \frac{px}{q} + \frac{pm}{qn}\\
&=qnx^2 + (mq + pn)x + pm.
\end{align}
This is quite different from the quadratic $x^2 + bx + c$.
Am I along the right path? What's the complete reasoning?

Notice that if $u = \frac{p}{q}$ and $v = \frac{m}{n}$ for some integers $p,q,m,n$ then you obtain the equation $$qn\left(x + \frac{p}{q}\right)\left(x + \frac{m}{n}\right) = qnx^2 + (mq + pn)x + pm.$$  In the expression $qnx^2 + (mq + pn)x + pm$ we can conclude that $qn, mq+pn$, and $pm$ are all integers.  If we use the labels
\begin{align}a &= qn\\
b&=mq+pn\\
c&=pm\\
u &= \frac{p}{q}\\
v &= \frac{m}{n}\end{align} then we again end up with $$(x-u)(x-v) = ax^2 + bx + c,$$ so in reality this form covers your case as well.  Generally, though, we don't require $a,b,c,u,v$ to be integers.
