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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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    @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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    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