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Normally I just guess a root and then smash one out in high degree functions, or complete squares or any other number of mathemagical tricks, but my textbook has decided to break numbers on me and I really don't get my roots here.

The question is as follows:

Find all zeros of the polynomial $f(x)=x^{2}+2x+2$

$1$. in $\mathbb{Z}_{3}$

$2$. in $\mathbb{Z}_{5}$

$3$. in $\mathbb{R}$


For $3$, My attempt in $\mathbb{R}$: we have $(x+1)^{2} -1 +2 =0$ thus $x=1\pm i$, so clearly we cannot factor over the reals.

For $2$, clearly $1$ is a root as well as everything that belongs to $[1] \in \mathbb{Z}_{5}$; clearly any even number is a bust ( even * even + even - odd can't equal $0$) so the other one must be $3$, but it doesn't work so $1$ is the only root but our function is of degree $2$ so $x-1$ must have multiplicity $2$? Am I fair in this assumption?

For $1$ tried $1,2,3$ none work no zeros.

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If all else fails, your finite fields are all small enough to just substitute all possible numbers explicitly and see if you get zero. Another way is to replace your coefficients with something something equivalent. I.e. $f(x)=x^2-3x+2=(x-1)(x-2)$ in $Z_5$ – fixedp Apr 2 '13 at 3:35
im assuming if i can guess a root i can use synthetic division to bash the equation to a nicer form which would imply since none of the others in say $( Z_{5})$ for this example work would it be wise to assume that i have a repeated root $(x-1)^{2}$ for part 2? – Faust7 Apr 2 '13 at 3:38
You have $(x-1)(x-2)$ in part 2, so the roots are 1 and 2 with multiplicity 1 each... – fixedp Apr 2 '13 at 3:51
Given that you don't have a repeated root in part 2, clearly it's not wise to assume so! Once you found $x=1$ as a root I would have done the division to find the other root explicitly; when you find it you can plug it in to confirm. – Steven Stadnicki Apr 2 '13 at 4:22
up vote 9 down vote accepted

You can (in each case) start by realizing that $$x^2+2x+2=0$$ if and only if $$(x+1)^2=-1$$ in whatever field you're working in. You've already noted that this doesn't work in $\Bbb R.$

In $\Bbb Z_3$, this becomes $$(x+1)^2=2,$$ but the only squares in $\Bbb Z_3$ are $0$ and $1,$ so this is impossible. In $\Bbb Z_5$, it becomes $$(x+1)^2=4.$$ Now, $2^2=3^2=4$, so $1$ and $2$ are the zeros of $f(x)$ in $\Bbb Z_5$.

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As none of the fields has characteristic $\,2\,$ you can use all we learned in high school about the quadratic roots formula, discriminant and stuff, so the question would be: is the quadratic's discriminant

$$\Delta:=b^2-4ac=2^2-4\cdot 1\cdot 2=-4$$

a square in our field? This is equivalent to ask whether $\,-1\,$ is a square in our field...and there are two among the ones your wrote down in which it is not but one in which it is, so...

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