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I'm studing polynomials; I have this exercise:

Find the irreducible factors of the polynomial $x^4-2x^2-3 \in \mathbb{Z}_5[x]$

I think in this way: I need to find root of the polynomial. A root of polynomial is a number such that the polynomials application $f(x) = x^4-2x^2-3=0$ that means $x^4-2x^2-3 \equiv 0 \pmod 5$, so $5|x^4-2x^2-3$. The only $x$ that makes this possible is $x=2$ (in fact $5|5$) and $x=3$ (in fact $5|60$). I know that this is a correct way, but how can I find roots if I'm in $\mathbb{Z}_{430}$, obviously I can't try this 430 times. Again: what if I'm in $\mathbb{R}[x]$?

Anyway, the next step is to divide the polynomial by $x-i$ where $i$ are my roots. So

$$\frac{x^4-2x^2-3}{x-2}= x^3+2x^2+2x+4$$ $$\frac{x^3+2x^2+2x+4}{x-3} = x^2+2$$

Since $x^2+2$ is irreducible, the factorization is $(x-2)(x-3)(x^2+2)$ is right?

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    Note that by setting $z = x^2$, you can simplify this to $z^2 - 2z - 3$. Since $2 \nmid p$, you can then apply the binominal formulas to bring this equation into the form $(z+a)^2 = b$ and solve this by finding a square root of $b$ in $\mathbb Z_p$. In a second step, solve $x^2 = z$ for the values of $z$ you find. This approach works for this special formula over all fields with characteristic $\neq 2$, and with small adaptations even in that case.2012-03-20
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    The high school factorization $(x^2-3)(x^2+1)$ is universal. Further factorization is field-dependent.2012-03-20

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