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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 $\,\sqrt 2\notin\Bbb Z_5\,$ –  DonAntonio Jul 15 '12 at 11:29
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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)? –  Per Manne Jul 15 '12 at 11:51
    
In English, not "grade" but "degree". –  GEdgar Jul 15 '12 at 12:46
    
@PerManne I'm sorry, but I'm afraid I don't understand the point you're making. –  haunted85 Jul 15 '12 at 12:52
    
@GEdgar I was in a rush and I didn't pay much attention to the language, I apologize. –  haunted85 Jul 15 '12 at 12:53

1 Answer 1

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? –  Gerry Myerson Jul 15 '12 at 12:21
    
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... –  DonAntonio Jul 15 '12 at 13:48
    
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. –  Gerry Myerson Jul 16 '12 at 10:43
    
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. –  DonAntonio Jul 16 '12 at 11:05
    
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. –  Gerry Myerson Jul 16 '12 at 13:28

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