Prove that x^3-9 is irreducible over the integers mod 31. Prove that x^3-9 is reducible over the integers mod 11. Any help in this would be greatly appreciated as I study for my final exam.
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Originally Posted by apalmer3 Prove that x^3-9 is irreducible over the integers mod 31 Prove that x^3-9 is reducible over the integers mod 11. Any help in this would be greatly appreciated as I study for my final exam. It is sufficient here to prove the polynomials have no zeros.
Okay! I see how that helps prove that it's irreducible over the integers mod 31... but how does that prove that it's reducible over integers mod 11? Thanks for the help!
Actually... I don't understand... For example: x^2+1 is reducible mod 5. It's equal to (x+2)(x+3). How can that be when (real) roots don't exist?
Originally Posted by apalmer3 but how does that prove that it's reducible over integers mod 11? Note $\displaystyle 4$ is a zero, thus, $\displaystyle (x-4)$ is a factor of $\displaystyle x^3 - 9$.
Originally Posted by apalmer3 Actually... I don't understand... For example: x^2+1 is reducible mod 5. It's equal to (x+2)(x+3). How can that be when (real) roots don't exist? That only works for polynomials over $\displaystyle \mathbb{R}$. You are in a different field, i.e. mod 5.
Originally Posted by ThePerfectHacker That only works for polynomials over $\displaystyle \mathbb{R}$. You are in a different field, i.e. mod 5. Okay. But (x+2)(x+3) = x^2+5x+6 In the integers mod 5, that equals x^2+0x+1 = x^2+1... see what I mean?
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