What are the irreducible polynomials in Z5[x]?

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Homework Statement


Factor f(x) = 3x4 + 2 into a product of irreducible polynomials in Z5[x]

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The Attempt at a Solution



I don't get it. I tried dividing it using the division logarithm, but then I can only get it to a point where it's like, 3(x-1)(..) <- polynomial of degree 3
Just simply plugging in values of Z5 doesn't seem to work..

I know there's some sort of trick to use.. I don't really understand how to factor f(x) using the different degrees. :S My textbook does a poor job of explaining it, without any concrete examples for me to go by, and I tried googling it but only saw an unanswered question.

If anyone can explain or direct me to some websites that explain how to do this, that would be great.
:)
 
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for clarity, how about constructing a table something like
x _ x^4 _ 3x^4 _ 3x^4 mod 5 _ (3x^4 +2)mod5
0 ...
1 ...
2 ...
..
 
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and fill in all the values? for each element in Z5?
 
well just to get a handle on how it all works in Z5, this latex formatting may help
<br /> \begin{matrix}<br /> x &amp; x^4 &amp; 3x^4 &amp; 3x^4 +2 &amp; (3x^4 +2)mod5\\<br /> 0 &amp; 0 &amp; 0 &amp; 2&amp;2 \\<br /> 1&amp; 1&amp; 3&amp;5 &amp; 0 \\<br /> ...&amp; &amp; &amp; &amp; &amp; \\<br /> \end{matrix}<br />
 
Got it thanks!
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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