Here's hoping I won't embarrass myself too much...

Since presumably your teacher already wrote the solution onto the board, I'll just talk about how I would arrive at a solution.

So first of all, it is a very common problem. We must have solved countless polynomial problems at school. So usually, seeing this kind of problem should already ring several bells.

So what I associate with polynomials (from school day) is

- zero points

- factorization

- quadratic equation

- "curve discussion" (don't know the english name for it, determining minima and maxima by calculating differentials)

Did you have the same associations? If not, then that is simply a matter of experience that would come with practice. Without that experience, naturally arriving at a solution would be hard.

It should be immediately obvious that we can factorize this polynomial without problems: simply because we know that we can factorize every second degree polynomial, thanks to the quadratic equation. This knowledge is basically "rote learning", except that you'll probably pick it up without decidedly doing rote learning. The polynomial presented here is already factorized into two second degree polynomials, so no problem there - do you agree so far? (Bonus points for seeing immediately that (x^2-a^2) = (x+a)(x-a) - this is a really common and basic formula that you should probably just "know". However, in principle you can also derive it from the quadratic equation, knowing this formula is just a shortcut).

So, are these zero points of use after all? Yes!!! We can simply check some values of the function between the zero points and determine if they are greater or less than zero. Since the function can only "switch sides" at the zero points, we then know it has to have the same sign (negative or positive) everywhere between those zero points (and same for outside of all the zero points).

Do you agree so far? If not, where is that you were stuck?

With that idea, let's just determine the zero points. From the (x+a)(x-a) part we already know that a and -a are zero points.

Using the quadratic formula on x^2 - 2x + a^|a| I get -1-sqrt(1-2^|a|) and -1+sqrt(1-2^|a|) - just ask if that is too fast, also be aware that I am very prone to make errors in such calculations.

From there on, I admit it is not much fun. We have to somehow order the zero points, so that we now when the sign could change. So we have to make case distinctions.

In fact, I haven't yet thought further than this, and now I am looking for ways to make it easier (I hate case distinctions). Laziness is the main motivator for maths, therefore I am not that much in favor of the rote learning approach...

So one step back: we have a products of two polynomials, poly1*poly2 with poly2 = x^2-a^2. That product is > 0 if poly1 > 0 and poly2 > 0 or poly1 < 0 and poly2 < 0 (if this is not obvious, please ask).

It is clear that poly2 < 0 for |x| < |a| and > 0 for |x| > |a|.

So let's look at poly1 = (x+1-sqrt(1-2^|a|))(x+1+sqrt(1-2^|a|)) (if my calculation was correct).

Since 1-sqrt(1-2^|a|) < 1+sqrt(1-2^|a|) we get

poly1 < 0 for -(1+sqrt(1-2^|a|)) < x < -(1-sqrt(1-2^|a|)) and > 0 for x < -(1+sqrt(1-2^|a|)) or x > -(1-sqrt(1-2^|a|))

So alltogether, the (intermediate) solution is

(-(1+sqrt(1-2^|a|)) < x < -(1-sqrt(1-2^|a|)) and |x| < |a|)
or (x < -(1+sqrt(1-2^|a|)) or x > -(1-sqrt(1-2^|a|))) and |x| > |a|

I have a feeling this should be simplified further, but I'll post it like this because it becomes a little bit intractable in the small comment box of HN. Maybe I'll find time to finish later.

Anyway, I probably made mistakes, and overlooked some much easier approach. The main point is that most of the steps are very common approaches, and I think/hope I demonstrated dividing the problem into smaller sub-problems.

One advice I would give for maths is to not be intimidated - you need a little bit of faith... If you fear the equations, you'll be too paralyzed to find a solution.

Also, I used to cheat a little bit on the problems in school: usually chances are high that the problem at hand has something to do with a recent thing the teacher has taught us. So it pays to look for ways to use the knowledge from the latest lessons.

Also, this problem seems comparatively ugly.

Sorry that this got rather complicated. Let me know how I can help better. I'll try to create a shorter summary of the steps involved. (Meanwhile, please post the shorter solution if there is one and you find it).