Factoring $x^5 + x^4 + x^3 + x^2 + x + 1$ without using $\frac{x^n - 1}{x-1}$?

Question

I was at a math team meet today and one of the problems was to factor $x^5 + x^4 + x^3 + x^2 + x + 1$. It also gave the hint that it decomposes into two trinomials and a binomial.

The solution they gave was based on the fact that $\frac{x^6 - 1}{x-1} = x^5 + x^4 + x^3 + x^2 + x + 1$ and from there the solution is pretty straightforward. However, I was not aware of that factorization. The only ones I have really learned are $x^2 - y^2 = (x-y)(x+y)$ and $x^3 \pm y^3 = (x \pm y)(x^2 \mp xy + y^2)$. Is there any other way I could have solved this factorization without using the ones they used?

According to the rational root theorem, its only rational roots can be $\pm1$. Indeed, $-1$ works. — Lucian, Mar 17 '14 at 23:59
You don't, but you look for those first, because the rational roots theorem gives the candidates easily. — Vedran Šego, Mar 18 '14 at 00:05
Oh perhaps the hint that one of the factors is a binomial implies one of the factors is an integer (hence rational)? If that's what they meant by the hint that is clever — user99185, Mar 18 '14 at 00:08
I don't recall saying that. But since we're dealing with a quintic, and since equations of degree $\geqslant5$ possess no general formula for roots, this is our only shot. Wouldn't you agree ? — Lucian, Mar 18 '14 at 00:10
@Lucian Sorry, I meant the question gave that information as a hint. — user99185, Mar 18 '14 at 00:12
@VedranŠego: Actually, I'm $\geqslant30$, not just $\geqslant5$. :-) — Lucian, Mar 18 '14 at 00:14
At OP, In the general case finding roots is hard, however cases like this invite inspection, ESPECIALLY this one as there are ... it's only got 1s. Look you can get -3 from the odd powers, +2 from the even (and >0) powers, +1 = 0, therefore -1 is a root, inspection! Does not help generally, helps specifically here! — Alec Teal, Mar 18 '14 at 00:20

score 10 · Answer 1 · edited Mar 18 '14 at 00:08

10

You can still factor this. Notice that

\begin{align*} x^5+x^4+x^3+x^2+x+1 &= x^3(x^2+x+1)+1(x^2+x+1) \\ &=(x^3+1)(x^2+x+1) \\ &=(x+1)(x^2-x+1)(x^2+x+1). \end{align*}

Now you can go one step further and show that both $(x^2-x+1)$ and $(x^2+x+1)$ are irreducible in the real numbers just by using the discriminant.

edited Mar 18 '14 at 00:08

Vedran Šego

11,372

answered Mar 18 '14 at 00:06

user60887

2,935

score 3 · Answer 2 · answered Mar 18 '14 at 00:02

3

Here's how you could have found that identity without having known it before hand.

First, notice that setting $x = -1$ gives $0$. Thus, $(x - (-1)) = x+1$ is a root.

Using polynomial long division, we find that we can reduce it to $(x^4 + x^2 + 1)(x+1)$

Now, what to do with the $x^4 + x^2 + 1$ term? Recall that $x^3 - 1 = (x-1)(x^2 + x + 1)$. If we plug $x = x^2$ into this identity, we find $x^6 - 1 = (x^2 - 1)(x^4 + x^2 + 1)$, so $x^4 + x^2 + 1 = \frac{x^6 - 1}{x^2 - 1}$.

We have now reduced it to $\frac{(x^6 - 1)(x+1)}{(x-1)(x+1)}$ = $\frac{x^6 - 1}{x-1}$ and, as you said, the rest is easy from here using the identities you listed in the OP.

answered Mar 18 '14 at 00:02

MT_

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Or one could notice that if we "shift" $x^5+\cdots+1$ by one place (read: multiply by $x$), we get $x^6+\cdots+x$ and the difference between the two is $x^6-1$, which looks quite nice for a factorization. – Vedran Šego Mar 18 '14 at 00:07
Yep. Though I would not think that that would be the most obvious thing to do if the only factoring one has encountered are sum/difference of two cubes/sqaures – MT_ Mar 18 '14 at 00:10
For me neither, but some students tend to be better with observing such patterns than with using existing formulas. – Vedran Šego Mar 18 '14 at 00:12

score 2 · Answer 3 · answered Mar 18 '14 at 00:35

You should be(come) aware of: $\frac{x^n-1}{x-1} = \sum_{k=0}^{n-1} x^k$

It's quite useful.

$$x^5+x^4+x^3+x^2+x+1 \\ = \frac{x^6-1}{x-1} \\ = \frac{(x^3-1)(x^3+1)}{x-1} \\ = (x^2+x+1)(x^3+1) \\ = (x^2+x+1)(x+1)(x^2-x+1)$$

The only other way would be to guess -1 as a root, because you have six terms in ascending polynomial sequence.

$$ x^5+x^4+x^3+x^2+x+1 = (x+1)(x^4+x^2+1)$$

Then split the resulting tetranomial, by solving: $$\exists a, b : x^4 + x^2 + 1 = (x^2+ax+1)(x^2+bx+1) \\ (x^2+a x + 1)(x^2+b x +1) = x^4 + (a+b) x^3 + (2+ab) x^2 + (a+b)x + 1 \\ \therefore a+b=0 \land ab=-1 \\ \therefore a = \pm 1, b=\mp 1 $$

And so: $$ x^5+x^4+x^3+x^2+x+1 = (x+1)(x^2+x+1)(x^2-x+1)$$

Factoring $x^5 + x^4 + x^3 + x^2 + x + 1$ without using $\frac{x^n - 1}{x-1}$?

3 Answers3