Show that $Q(x):=200x^3-200x^2+200x+100$ is an irreducible polynomial over the field $\mathbb{Q}$ of rational numbers.

Question

I'm trying to use Eisenstein's Criterion to prove this but I think I don't quite understand the theorem maybe. For example if I let $p=2$ then don't both 2 and 4 divide each coefficient?

Just use Eisenstein directly on the reciprocal, see https://math.stackexchange.com/questions/963477/prove-that-the-polynomial-xnf1-x-with-reverted-coefficients-is-also-irreduc — Sil, Apr 23 '20 at 09:55

score 6 · Answer 1 · edited Apr 22 '20 at 23:51

6

First, you can factor out $100$. Let $$P(x)=2x^3 - 2x^2 + 2x + 1.$$ This is an associate of $Q(x)$, so $Q$ is irreducible if and only if $P$ is irreducible.

Now, either apply the “backwards Eisenstein’s Criterion” to $P$, or if you don’t know the backwards Eisenstein, then either prove it (same proof; check your understanding of the original criterion by proving this one), or look at $$R(x) = x^3P\left(\frac{1}{x}\right)= 2 - 2x + 2x^2 + x^3.$$ Show $R$ is irreducible, and conclude that $P$ must also be irreducible.

edited Apr 22 '20 at 23:51

Calvin Lin

68,864

answered Apr 22 '20 at 23:35

Arturo Magidin

398,050

1

@OP see here for more on applying Eisenstein to the reversed ("reciprocal") polynomial. – Bill Dubuque Apr 22 '20 at 23:44

score 5 · Answer 2 · answered Apr 23 '20 at 00:13

5

You don't need Eisenstein Criterion, a cubic polynomial is irreducible over $\mathbb Q \iff$ it doesn't have rational roots. Since $P(x) = 100(2x^3-2x^2+2x+1)$, you just have to check $\pm1$ and $\pm \frac{1}{2}$.

answered Apr 23 '20 at 00:13

Sam

1,381

Show that $Q(x):=200x^3-200x^2+200x+100$ is an irreducible polynomial over the field $\mathbb{Q}$ of rational numbers.

2 Answers2