Twelfth grader Abbey wants some help with the following: "Factor x⁶ +2x⁵ - 4x⁴ - 8x³ + x² - 4."

Well, Abbey, if you've read our unit on factoring higher degree polynomials, and especially our sections on grouping terms and aggressive grouping, you probably realize that a good way to attack this problem is to try grouping the terms. Hopefully, you tried something along those lines. Here's one method (I'll get you started on another method at the end of this post):

(x⁶ + 2x⁵) - (4x⁴ + 8x³) + (x² - 4).

That looks promising, since every group can be factored:

x⁵(x + 2) - 4x³(x + 2) + (x - 2)(x + 2).

Great! every group has an (x + 2) term in it, so we can factor that out:

(x + 2)(x⁵ - 4x³ + x - 2)

I'm going to attempt grouping on that second polynomial:

(x + 2)[(x⁵ - 4x³) + (x - 2)]

(x + 2)[x³(x² - 4) + (x - 2)]

(x + 2)[x³(x - 2)(x + 2) + (x - 2)]

And once again we have a common factor in our groups - this time it's (x - 2).

(x + 2)(x - 2)[x³(x + 2) + 1]

(x + 2)(x - 2)(x⁴ + 2x³ + 1)

Now we have something that can't easily be grouped, because it has 3 terms. So I decided to try something crazy: I added in x² and subtracted it back out again. Why? Take a look:

(x + 2)(x - 2)(x⁴ + 2x³ + x² + 1 - x²)

(x + 2)(x - 2)[(x⁴ + 2x³ + x²) - (x²- 1)]

(x + 2)(x - 2)[x²(x² +2x + 1) - (x + 1)(x - 1)]

(x + 2)(x - 2)[x²(x + 1)² - (x + 1)(x - 1)]

My groups each contain an (x + 1), so I can factor that out:

(x + 2)(x - 2)(x + 1)[x²(x + 1) - (x - 1)]

(x + 2)(x - 2)(x + 1)(x³ +x² - x + 1)

At this point, I tried a couple different groupings, and nothing seemed to work to factor (x³ +x² - x + 1). That doesn't mean it's not factorable, but I got a little suspicious, so I pulled out the Rational Root Theorem, which tells me that if that thing has any rational zeroes, they have to be either 1 or -1. Since neither of those are zeroes of the polynomial, that means it's unfactorable over the rationals. Therefore, the final answer is:

(x + 2)(x - 2)(x + 1)(x³ +x² - x + 1)

Just a couple notes on this, before I'm done:

1. I enjoy the process of trying to do grouping (and what I call "aggressive grouping") on polynomials in order to factor them. I think it's a good mental exercise. Nevertheless, I should point out that according to the Rational Root Theorem, if this polynomial has any rational zeroes, they are any of the following: -1, 1, -2, 2, -4, 4. Trying out those to see which work, and then using Synthetic Division to divide out the factors will guarantee you a solution, even if you can't figure out groupings to do.
2. There's another way to start off the problem (which is actually what I did the first time I solved the problem), which involves rearranging the terms before grouping: (x⁶  - 4x⁴) + (2x⁵- 8x³) + (x² - 4). Grouping this way has the advantage of pulling out (x + 2)(x - 2) all in one step, and going straight to a 4th degree polynomial. It's nice to see that the problem can be done in multiple ways, don't you think?