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 $x^2 + bx + c$ factors then by quadratic equation it must factor to
$(x - \frac {-b+ \sqrt{b^2 - 4c}}2) (x - \frac {-b- \sqrt{b^2 - 4c}}2)$ and this factors if and only if $b^2 - 4c$ is a perfect square (note: if $b$ is even/odd then $b^2 -4c$ is even/odd so $-b\pm \sqrt{b^2-4c}$ will be even so either $\frac {-b \pm \sqrt{b^2 -4c}}2$ is an integer or $x^2 + bx + c$ is not factorable.
But if $b^2-4c$ is a perfect square then 
$x^2 -bx +c = (x -\frac {b+\sqrt{(-b)^2-4c}}2)(x-\frac {b+\sqrt{(-b)^2 -4c}}2)$ is also factorable.

Although a much easier idea is by user744868  in the comments.
If $f(x) = x^2 + bx + c = (x+r)(x+s)$ then $f(-x) = x^2 - bx + c= (-x+r)(-x+s)=(x-r)(x-s)$.
