Factoring $x^2 - 2 + \frac1{x^2}$ by steps

$$x^2 - 2 + \frac1{x^2}$$ 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.

• Please use MathJax to format your equations.
– Gary
Commented Feb 6 at 23:27
• Hi, welcome to Math SE. Would you have also been stuck on $x^4-2x^2+1$? What about $y^2-2y+1$?
– J.G.
Commented Feb 6 at 23:31
• @J.G. has a nice comment - if you can't see this immediately (that takes time when you have not seen this kind of thing before) then put everything over the common denominator $x^2$ so you are dealing with polynomials in numerator and denominator. And you know what to do with polynomials. Commented Feb 6 at 23:36
• The trick to realize is that $2= 2x\cdot x$. But even if I didn't see that, I'd factor out the $\frac 1{x^2}$ to get it all as a polynomial with positive terms. $x^2 -2 + \frac 1{x^2} = \frac 1{x^2}(x^4 -2x^2 + 1)$ and that is apparent what to do.... or we could try to figure $(x +a\frac 1x + c)(x+b\frac 1x + d)=x^2 +a+cx + b+ab\frac 1{x^2} + cb\frac 1x + dx + ad\frac 1x + ad=x^2 + (c+d)x +(b+a+cd) +(ad+cb)\frac 1x +ab\frac 1x^2$ so $c+d=0;ad+cb=0;ab=1;b+a+cd = -2$. Commented Feb 7 at 2:23
• Seeing this is instant with experience. As much as it’s an algorithm as the answers describe, it’s also memorization and recognition. With time you will see that $x^{2m}+2p$ is calling for $\frac{p}{x^{-m}}$ to complete the square, if you’re trying to get out of a raised power. Commented Feb 7 at 3:53

2 Answers

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.

• I was working through @J.G 's clues and got stuck at precisely here, "Now plug z=x2". I forgot that since we multiplied everything by x squared in the beginning we had to undo that at the end. Thanks for the guidance! I have to learn math jacks now. Thanks to whoever reformatted my question. Commented Feb 6 at 23:56

$$x^2 -2+\frac{1}{x} = \frac{x^4-2x^2+1}{x^2}$$ Let $$u=x^2$$, we get $$u^2-2u+1\implies (u-1)^2$$ Finally $$\frac{(u-1)^2}{x^2} = \frac{(x^2-1)^2}{x^2} = (x^2-1)^2x^{-2} = (x^2-1)x^{-1}(x^2-1)x^{-1}.$$

• Isn’t this exactly what’s already in the first answer? Commented Feb 7 at 0:55