# Factoring polynomial over rings

1. Nov 28, 2011

### llstelle

1. The problem statement, all variables and given/known data

Factor x^3+3x+6 over Z5, Z10, Z, Q and ℝ.

2. The attempt at a solution

For Z5, I have the roots of x^3+3x+1 in Z5, x=2=-3 and x=3=-4, so x+3 and x+4 are factors. By long division of x^3+3x+1, it is found that x+3 is a repeated factor, so x^3+3x+6=(x+4)(x+3)^2.

For Z and Q, Eisenhart's criterion is satisfied with 3, hence x^3+3x+6 is irreducible over Q, and hence Z.

For Z10 and ℝ... I don't really get how to do this problem. Is there a consistent way to work out the factors for Z10? It seems like a lot of effort. Also, I noticed that x^3+3x+6 has one real root over ℝ, and so I can factor it neatly to a linear factor (with real root) and a quadratic factor with two complex roots, but I only managed to get these roots with a computational method - and they have pretty long but explicit expressions, so again, my question: is there a proper way to work out the roots by hand here?

Thanks!

2. Nov 28, 2011

### I like Serena

Welcome to PF, llstelle!

First off, x=3 is not a root in Z5, but x=1 is.

For Z10 you only need 1 root. Is perhaps x=1 or x=2 also a root in Z10?

For R, I think you're stuck with the complicated solutions.
But here's a general method that gives the answer fairly easily:

Starting with $x^3+3x+6=0$.

Substitute x=y+z, meaning you have a free choice for either y or z.
So $(y+z)^3+3(y+z)+6=0$

$\Rightarrow (y^3+z^3+3y^2z+3yz^2) + 3(y+z)+6=0$

$\Rightarrow y^3+z^3+3(yz+1)(y+z)+6=0$

Choose z such that $yz+1=0$, or $z=-{1 \over y}$
Then: $y^3 - {1 \over y^3} + 6 = 0$
$\Rightarrow (y^3)^2 + 6(y^3) - 1 = 0$

Solve as a quadratic equation and back substituting z gives:
$$x=y+z=\sqrt[3]{-3 + \sqrt{10}} - {1 \over \sqrt[3]{-3 + \sqrt{10}}}$$
or
$$x=y+z=-\sqrt[3]{3 + \sqrt{10}} + {1 \over \sqrt[3]{3 + \sqrt{10}}}$$

Note that both solutions are the same root.

3. Nov 28, 2011

### llstelle

Hello! Oh oops, that was a typo - yes I have x=1 as a root of the equation for Z5. And thanks for the general solution for ℝ! I've never done that before. I've been taught to test numbers until I find a root, then divide the cubic equation by the linear factor to find a quadratic factor, and use the general solution for a quadratic equation for the quadratic factor.. ): This is so much neater!

OK, I tried each of them for Z10 and I have x=1, 2, 6 or 7 satisfying f(x)=0. If I write (x+9)(x+8)(x+4)(x+3), I get a fourth order polynomial whose fourth order term does not vanish in Z10... How does this happen? Something looks wrong to me.

(:

4. Nov 28, 2011

### I like Serena

A polynomial of degree n can have more than n roots in Zn if n is not prime.
For instance, $x^2-1$ has 8 roots in Z24: $\pm 1, \pm 5, \pm 7, \pm 11$.

Last edited: Nov 28, 2011
5. Nov 28, 2011

### Dick

And would probably indicate that x^3+3*x+6 doesn't have a unique factorization. Z10 has zero divisors. You can probably find several factorizations.