0

I asked this question. Now I am trying to factor the same polynomial, $f(x)=x^5+2x^4+3x^3+5$, in $\mathbb{Z}_3[x]$.

So, we have that $f(x)=x^5+2x^4+2$ in $\mathbb{Z}_3[x]$. Like in my previous question I start by considering the roots, which gives $f(0)=2, f(1)=2$, but $f(2)=32+32+2=66=0$. Hence the polynomial has a root at $2$. This would mean $f(x)=(x+2)g(x)$ for some $g(x)\in \mathbb{Z}_3[x]$. I could not find such $g(x)$. Instead I went to the an online calculator that shows the answer, namely $(x^3+2x+2)(x+1)^2$ (which is correct).

Furthermore, $(x+1)^2=x^2+2x+1$ has a root at $2$. Why trying to factor using $(x+2)$ does not work, and how could one have come up with the $(x+1)^2$ term knowing that the polynomial has a root at $2$?

Basically, I have the answer, but I'd like to know how to derive it.

user26857
  • 52,094
Schach21
  • 700

1 Answers1

2

If a polynomial has a root at 2, then $(x-2)$ is a factor. Modulo 3, $(x-2)$ becomes $(x+1).$ This is how you derive.

  • Oh, of course! What about $(x+1)^2$ why the square? Sorry if this is too trivial. – Schach21 May 11 '21 at 21:23
  • 1
    Because $2$ is a zero of multiplicity $2$. One can see this as follows: If you divide the given polynomial by $(x+1)$ you get $(x^4 + x^3+ 2x^2 - 2x + 2)$. However, inserting $2=-1$ yields $1 + (-1) + 2 - (-2) + 2 = 6 = 0$, therefore $(x+1)$ is a factor of this polynomial as well and therefore $(x+1)^2$ is a factor of the degree $5$ polynomial you started with – LegNaiB May 11 '21 at 21:31
  • 1
    Another way to figure this out is to look at the "formal derivative" if you're familiar with it (https://en.wikipedia.org/wiki/Formal_derivative#Application_to_finding_repeated_factors). Here, the formal derivative is $5x^4 +8x^3 + 9x^2$. Plugging in $x = 2$ gives $180 \equiv 0 \mod 3$. A formal derivative of $0$ indicates a multiple root. – Osama Ghani May 11 '21 at 21:33
  • 2
    Just mentioning that the formal derivative - as well as the original function - can be represented with much smaller coefficients: $5x^4+8x^3+9x^2 = -x^4-x^3$. Using this evaluation is much easier, especially if $-1$ instead of $2$ is inserted. I always use this trick to avoid using the calculator for that. – LegNaiB May 11 '21 at 21:40