How can I prove that if $x^2+bx+c$ is factorable, then $x^2-bx+c$ is also factorable? I want to prove that if $x^2+bx+c$ is factorable, then $x^2-bx+c$ is also factorable(by factorable I mean that it can be expressed with the product of $2$ binomials $(x+y)(x+z)$, where $y,z\in\mathbb Z$). Also, $b,c\in\mathbb Z$. It seems to be true in all the quadratic expressions I have tested, but I'm not too sure how to prove such a thing. Can someone help me prove this or provide a counterexample?
MY ATTEMPT:
$x^2+bx+c$ is factorable, so I can write it as $(x+y)(x+z)$ for some integer values $y$ and $z$. $y+z=b$, and $yz=c$.
I will initially assume that $x^2-bx+c$ is factorable and then see what I get:
$x^2-bx+c$ is factorable, so I can write it as $(x+p)(x+q)$ for $p,q\in\mathbb Z$. $p+q=-b$, and $pq=c.$ 
I have to somehow prove that $p$ and $q$ are integers but I'm not too sure how. Any advice would be greatly appreciated.

If
$x^2+ax+b
=(x-u)(x-v)
$
then
$x^2-ax+b
=(x+u)(x+v)
$
since
$a = -u-v
$
and
$b = uv$.
