2

I wish to determine that $f(x) = x^4 - 6x^2 - 8x + 3$ is irreducible in $\mathbb{Q}[x]$ by substituting $x + m$ for $x$ and then choosing $m$ appropriately.

$\textit{Work so far:}$ By rational root test, we have $\pm 1, \pm 3$ $$f(1) = 1-6-8+3 = -10$$ $$f(-1) = 1-6+8 +3 = 6$$ $$f(3) = 3^4 - 6\cdot 3^2 - 8\cdot 3 + 3 = 6$$ $$f(-3) = (-3)^4 - 6\cdot (-3)^2 - 8(-3) + 3 = 54$$ none equal zero.

I am not sure where to go from here. I know that since $f$ has degree $4\ge 2$ then if $f$ has a root then it is irreducible. I am not sure what to do in order to apply Eisenstein's criterion, or if there is another method to use (Gauss' Theorem). Can someone please help me with this?

rudinsimons12
  • 1,355
  • 2
    The general Eisenstein method is explained at length in the 2nd linked dupe, i.e. by this answer we see that $,f(x) \equiv (x+1)^4\pmod 2,$ so $,f(x-1),$ is Eisenstein for $,p=2\ $ There are many dupes showing how to apply Eisenstein via such a shift. – Bill Dubuque Nov 19 '21 at 19:57

0 Answers0