$$x^2 - 2 + \frac1{x^2}$$ factors out to $$\left(x-\frac1x\right)^2$$ I know this because the answer was given to me in a video. If the answer wasn't given to me, I would have been stuck. I'm not even sure how to plug this in quadratic formula because I don't know what the value for $c$ should be.
$$\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$
What am I failing to grasp? I bet it is something simple I forgot.
Expanding the @Mark-Bennet comment
$$ x^2 - 2 + \frac{1}{x^2} = \frac{x^4}{x^2} - \frac{2x^2}{x^2} + \frac{1}{x^2}\\ = \frac{1}{x^2} ( x^4 - 2x^2 + 1)\\ $$
Now you have to factor the ordinary polynomial $x^4 - 2x^2 + 1$.
See that this only depends on $z \equiv x^2$ so you can write it as $z^2 - 2z + 1$.
You can factor this as $(z-1)^2$ by recognizing it from memory, but if you don't see that the quadratic formula will tell you $z^2 - 2z + 1$ has the root $1$ with multiplicity $2$ so you get $(z-1)^2$ from that.
Now plug $z=x^2$ back in and put it back with the $\frac{1}{x^2}$.
$$ \frac{1}{x^2} ( x^2 - 1)^2\\ $$
That is the product of two things squared, so you can just take the product before squaring.
$$ (\frac{1}{x} x^2 - \frac{1}{x} 1)^2 = (x-\frac{1}{x})^2 $$
As you gain practice with more and more, you will be able to fast-forward through steps and maybe get to a point where you can recognize it on sight.