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In ''Z''/15''Z'', (''x''+1) × (''x''+2) = ''x''<sup>2</sup> + 3''x'' + 2, and (''x''+7) × (''x''+11) is also ''x''<sup>2</sup> + 3''x'' + 2, and all of the polynomials ''x''+1, ''x''+2, ''x''+7 and ''x''+11 are irreducible (since all of they have degree 1), thus ''x''<sup>2</sup> + 3''x'' + 2 have two factorizations to irreducible polynomials in ''Z''/15''Z''. Besides, in ''Z''/15''Z'', ''x''<sup>2</sup> + 2 is irreducible, since −2 is not a quadratic residue mod 15.
:This article is about factorization ''over a finite field''. Thus it does not applies to {{math|''Z''/15''Z''}}, which is not a field. [[Chinese remainder theorem]] allows deducing the factorization(s) of a polynomial over {{math|''Z''/15''Z''}} from the factorizations over {{math|''Z''/3''Z''}} and {{math|''Z''/5''Z''}}. [[User:D.Lazard|D.Lazard]] ([[User talk:D.Lazard|talk]]) 19:56, 13 August 2018 (UTC)
== Trying to understand how any of these methods would work for f(x) with coefficients in GF(p^r) ==
Consider GF(2^16) based on primitive polynomial x^16 + x^12 + x^3 + x + 1. Find the factors of f(x) = x^32 + x^22 + x^2 + x + 1, where the coefficient values are 0 and 1, but are elements of GF(2^16). There are 16 quadratic factors, shown in hex (found by brute force search):
<pre>
x^2 + 04c4 x + 118d
x^2 + 09ad x + 1cec
x^2 + 0e38 x + 1cb7
x^2 + 16b6 x + 1dbc
x^2 + 173b x + 0cf9
x^2 + 1c89 x + 1cf0
x^2 + 40ab x + 4be1
x^2 + 524f x + 0a76
x^2 + 5e62 x + 0716
x^2 + 5eec x + 0a37
x^2 + b67f x + 4188
x^2 + be6f x + fbf0
x^2 + e079 x + 17d4
x^2 + effe x + ed71
x^2 + f62b x + 07d5
x^2 + fd83 x + 17dd
</pre>
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