# Factor poly in Z_5[x]

• Dec 3rd 2012, 09:12 PM
jzellt
Factor poly in Z_5[x]
Factor 2x^3 + 4x + 1 in Z_5[x] (modulo 5)

I can't figure this out. Can someone show please? Thanks!
• Dec 3rd 2012, 10:01 PM
topsquark
Re: Factor poly in Z_5[x]
Quote:

Originally Posted by jzellt
Factor 2x^3 + 4x + 1 in Z_5[x] (modulo 5)

I can't figure this out. Can someone show please? Thanks!

I'm sure there's a more general way to do this, but as we're talking about $\displaystyle \mathbb{Z}_5$ there are only 5 possibilities so it's not that hard to do a "chug and plug." If you want a bit more rigor, then note that $\displaystyle 1 \equiv -4$:
$\displaystyle 2x^3 + 4x + 1 \equiv 2x^3 + 4x - 4 \equiv 0$

Since 5 is a prime there is a unique inverse of 2 so we can multiply both sides by 3 and get
$\displaystyle x^3 + 2x - 2 \equiv 0$

The last two terms are even, so thus $\displaystyle x^3$ needs to be as well. Thus the only possible values for x are 0, 2, and 4.

-Dan
• Dec 4th 2012, 05:03 AM
Deveno
Re: Factor poly in Z_5[x]
Quote:

Originally Posted by topsquark
The last two terms are even, so thus $\displaystyle x^3$ needs to be as well. Thus the only possible values for x are 0, 2, and 4.

-Dan

everything up to here is true. but "even" has no meaning in Z5: 1 = 2+2+2. furthermore "0" in Z5 might mean some ODD multiple of 5 (as an ordinary integer).

however, it IS indeed true that: (2)(23) + (4)(2) + 1 = 0 (mod 5) and 2(43) + (4)(4) + 1 = 0 (mod 5).

i find that using a substitution like 1 = -4 (mod 5) can get confusing...it leads one to use properties of integers that DO NOT HOLD in Z5.

rather, multiply the original equation by 3:

3(2x3 + 4x + 1) = x3 + 2x + 3 = 0 (in Z5) <--this is a monic polynomial, so we're looking for monic factors.

we have two roots in Z5, so it must factor completely over Z5, so one of these (2 or 4) is a repeated root. we COULD find the repeated root by long division but meh...i'm lazy.

so let's take the derivative! we get:

3x2 + 2, and if we set this equal to 0 and multiply by 2, we get:

x2 + 4 = 0 <---this has the roots x = 1 and x = 4, the shared root with x3 + 2x + 3 is x = 4, so 4 is our repeated root.

thus:

x3 + 2x + 3 = (x + 1)2(x + 3) (i am writing x + 1 instead of x - 4, and x + 3 instead of x - 2), so:

2x3 + 4x + 1 = 2(x + 1)2(x + 3), as can be verified by multiplying the factors (you could also write this as 2(x - 4)2(x - 2), if you like seeing the roots in your factors).
• Dec 4th 2012, 09:54 AM
topsquark
Re: Factor poly in Z_5[x]
@Devano: Yeah, I saw that was wrong over lunch today. Thanks for the catch! :)

-Dan