AGREE

So true

Yess Yess

gotta love line wrapping

It's so true. 1/10 math videos are worthwhile.

Yuppp it is really

Hi

OH MY GOD YESSS

a*x^m + c, where m is relatively prime to (p-1) should work. i wonder if there are other cases

@@astrophilip Hmm... Indeed it seems to generate different results for each `x`. However, some arrangements of those results seem to be unreachable :/ Let's try the simplest one, for p=5, so p-1=4, and there's only one number coprime to it: 3, so let m=3, and then a·x³+c is our polynomial. First, let's check different a's in columns: x | x³ 2·x³ 3·x³ 4·x³ 0 | 0 0 0 0 1 | 1 2 3 4 2 | 3 1 4 2 3 | 2 4 1 3 4 | 4 3 2 1 For each of these columns, we can shift it by the offset c around the modulus and get 4 other such tables, or 4·5=20 different arrangements of the numbers from 0 to 4. But the number of all POSSIBLE arrangements is of course 5!=120, so we're missing 100 of them :q Even if we suppose that we only use the numbers 1..4 for our x, and replace 0 with whatever number is missing in the +c tables (so that the results were also in the 1..4 range), it's still only 20 out of 4!=24 possible arrangements. Here are the ones we're missing for 4-element sets: 1234, 2413, 3142, 4321 The 1234 can be accounted for if we allow the "identity" function x¹, but what about the other three? *Edit:* Nevermind, it seems that a·x¹ does the job, since: 1,2,3,4 ·2 = 2,4,1,3 1,2,3,4 ·3 = 3,1,4,2 1,2,3,4 ·4 = 4,3,2,1 Now I need to check if it still works for bigger moduli...

Eisenstein's Criterion cannot tell you that a polynomial is *reducible,* only that it is irreducible. If the criterion fails to apply, then it just tells you nothing; that's not a "false positive."

@@HamblinMath Yes, I understand that. I misspoke and should have said a 'false negative'. Clearly the polynomial is irreducible but it was not caught by either test. Why did the tests fail?

@@brandonk9299 There's not really a reason why the test failed. There isn't one perfect way of telling whether a polynomial is irreducible. That's why we have the several different methods in this video.

Correct, the criterion as I've stated it only applies in Q for primes in Z.

You are correct that 1 is a root of x^2+2, and thus x^2+2 must be reducible. In fact, we know that x - 1 (which is equal to x+2 in Z_3) must be a factor of x^2+2. Doing polynomial long division shows us that x^2+2 = (x+2)(x+1).

@@HamblinMath Oh, I have not realised these! Thank you very much for a quick response! By the way, perfect video.

The number "-1" doesn't really exist in Z_5, so if we're going to talk about the polynomial "4x^3 + x^2 - x + 3" in Z_5[x], we need to rewrite its coefficients in Z_5.

@@HamblinMath ahhh I see, so if we only need consider the polynomial in Q[X], and it is of degree 2 or 3, it is enough to show using the rational roots test? In essence, if we need to show its irreducible in Q[X] and it’s of a higher degree than 3, THEN I should try the reduction mod P test? Thanks for the swift response by the way you did a great job at explaining in this video:)

@@elliotbradfield3234 There's no surefire way to know which tool(s) will work for any given polynomial. The problem with reducing mod p for a polynomial with degree 4 or higher is that you'd still need to consider the possibility that it has a degree 2 factor.

No, if you can't find a prime for which Eisenstein's criterion applies, you cannot conclude anything about the polynomial. For example, neither x^2 + 1 and x^2 - 1 have a prime that works for Eisenstein's criterion, and one is irreducible and the other is reducible.

James Hamblin thank you so much!

if we choose degree three, there are more cases to consider and hence clumsy. namely, x^3+x^2+x+1 x^3+x^2+x x^3+x^2+1 x^3+x+1 x^3+x x^3+1 x^3

It has to have a factor of a combination of (1 and 3) or (2 and 2). If you're going to prove either, you only need to prove one of the factors. The 1 and 3 combination can't exist if there is no 1. Thus, to save yourself the time, you only need to prove that 1 or 2 can't exist, and leave 3 alone. You can use 3, but you're just creating more work for yourself for the same outcome.

If x=a is a root of your polynomial, then x-a is a factor. If you can't see how to factor it, try polynomial long division.