How can I prove that if $x^2+bx+c$ is factorable, then $x^2-bx+c$ is also factorable?

Question

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.

Answer 1

y+z=b , and yz=c

Its quite simple from ths step,

Put the value of b and c in your next equation.

Then you can easily factorize it and since y and z are integers, then p and q of the next equation will also be integers.

Answer 2

If

$x^2 + bx + c = (x - p)(x - q), \tag 1$

then since

$(x - p)(x - q) = x^2 - (p + q)x + pq, \tag 2$

we find that

$b = -(p + q), \tag 3$

$c = pq; \tag 4$

now reverse the signs of $p$ and $q$, and compute

$(x + p)(x + q) = x^2 + (p + q)x + pq = x^2 - bx + c, \tag 5$

which shows that $x^2 - bx + c$ is factorable provided $x^2 + bx + c$ is.

Note that, replacing $b$ with $-b$, we see that $x^2 + bx + c$ if factorable provided $x^2 - bx + c$ is.

It is worth observing, I think, that this result holds for

$x^2 + bx + c \in R[x], \tag 5$

where $R$ is any commutative ring; indeed, it may be possible to extend the coefficient ring even further; certainly something to ponder.

Answer 3

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 $f(-x) = x^2 - bx + c= (-x+r)(-x+s)=(x-r)(x-s)$.

Answer 4

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