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Find a degree 5 polynomial $f \in \mathbb Z_5[x]$ so that it has exactly 4 distinct roots and factorize it as product of irreducible factors.

I'm really struggling in finding such polynomial, so basically I need to find an $f$ which has a $c$ root that doesn't belong to $\mathbb Z_5$. So does this polynomial is the one I am looking for? If not is there any way I could get to such $f$?

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

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    Your example is *not* a pol. in $\,\Bbb Z_5[x]\,$ since $\,\sqrt 2\notin\Bbb Z_5\,$2012-07-15
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    Is it clear that multiple roots are disallowed? Can't you say that, e.g., $x^3-x^2$ has two distinct roots (and three when counting with multiplicity)?2012-07-15
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    In English, not "grade" but "degree".2012-07-15
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    @PerManne I'm sorry, but I'm afraid I don't understand the point you're making.2012-07-15
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    @GEdgar I was in a rush and I didn't pay much attention to the language, I apologize.2012-07-15
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    @haunted85 The example $x^3-x^2$ was supposed to be a little simpler than your case. The factorization is $x^3-x^2=x\cdot x\cdot(x-1)$, so it is a third degree polynomial with two distinct roots given by $x=0$ and $x=1$. A similar expression will give you a fifth degree polynomial with four distinct roots.2012-07-15

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What about the following examples?:

$$\,x^5-x^4-x+1$$ $$x^5+x^4-x-1$$

$$x^5+2x^4-x-2$$

Can you guess from what definite example are the above polynomials taken? If you can then factoring will be pretty expedite.

All the polynomials above have roots in $\,\Bbb F_5:=\Bbb Z/5\Bbb Z\,$ itself. If you want an example with a root out of this field, you can try $$x^5-x^4+x^3-x^2-2x+2$$

What's the idea to build the above example? Take an element in $\,\Bbb F_5\,$ which is not a quadratic residue (i.e., has not square root there), say $\,3\in\Bbb F_5\,$, then form its minimal polynomial $\,x^2-3=x^2+2\in\Bbb F_5[x]\,$ , and now just multiply this by a cubic with two different roots in $\,\Bbb F_5\,$ , say $\,x^2(x-1)\,\,,\,(x-1)(x^2-1)\,$ , etc.

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    That last construction will give you a polynomial with two distinct roots in the field, whereas OP wants four, no?2012-07-15
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    If you meant the pol. $\,x^5-x^4+x^3-x^2-2x+2=(x^2+2)(x-1)^2(x+1)\,$ it has exactly 4 distinct roots:$\,\pm\sqrt{- 2}\,,\,\pm 1\,$ , if I didn't commit a dumb mistake. If you meant the construction with $\,(x-1)(x^2-1)=(x-1)^2(x+1)\,$ we get two ddiferent roots here and the other two from the irreducible quadratic gives us 4 roots...2012-07-15
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    it's a question of differing interpretations of what OP wants. I took the question to be asking for 4 roots in ${\bf F}_5$, as the example OP gives has 5 roots, 4 of which are in ${\bf F}_5$. You take it to mean 4 distinct roots total, including ones outside ${\bf F}_5$. Since I can't be certain what OP intended, I withdraw my objection to your answer.2012-07-16
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    Ok, I think I get your point, @GerryMyerson. Nevertheless, if the OP really wanted a quintic with *only* 4 roots in $\,\Bbb F_5\,$ then it can't be done, as the fifth root *not* in $\,\Bbb F_5\,$ would add a linear factor to the quintic that is not in $\,\Bbb F_5[x]\,$ and thus the polynomial wouldn't be factored over $\,\Bbb F_5\,$...Anyway, in my examples I think both cases are covered So he can choose.2012-07-16
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    OP wants a quintic with only 4 *distinct* roots, and (if I have the correct interpretation of the question) Per Manne's comments point the way to a solution. It would be nice of OP to join us.2012-07-16
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    @GerryMyerson I would love to jump in and have a say but the thing is I don't understand much of what's goin' on here. I have asked such a question because I am very very confused about this matter, I was hoping to clear some air. In any case, you are right, your interpretation is correct I requested a degree 5 polynomial with four _distinct_ roots.2012-07-18
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    Well, a quintic with *only* 4 distinct roots must have exactly one double root (in some extension field), so I think my examples already covered all the possibilities...2012-07-18
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    haunted, Per Manne already hinted at a solution in the comments. To be explicit, doesn't $x^2(x-1)(x-2)(x-3)$ meet your needs? It has degree 5, it has exactly 4 distinct roots, and it is already factored into irreducibles. If this isn't what you want, please clarify.2012-07-18
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    @GerryMyerson yes $x^2(x-1)(x-2)(x-3)$ does do the trick. Now I am wondering if there is a linear factor so that the root does not belong to $\mathbb Z_5$. In other words is it possibile finding $f = x(x-a)(x-b)(x-c)(x-d)$ where at least one of the roots is not in $\mathbb Z_5$?2012-07-19
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    haunted, is this not covered in DonAntonio's answer? $x^2-3$ has no root in the field. If you still want 4 distinct roots, and you don't need them all to be in the field, you can use $x^2(x-1)(x^2-3)$.2012-07-19