Question.
$f(x)=x^5+4x^4+2x^3+3x^2+-x+5$ is irreducible over $\Bbb Q$?
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I know that use the prime number $p$ to $\bar f_{p}(x)$.
let $p=2$. then $\bar f_{2}(x)=x^5+x^2-x+\bar 1$.
but $x=1$, then $\bar f_{2}(1)=0$.
So i don't know about its state.
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Moreover, It can't use Eisenstein's criterion'.
Do i use the 'Method of Undetermined Coefficients'?
By Gauss's lemma, it suffices to prove that $p(x)$ is irreducible in $\mathbb{Z}[x]$. So
case 1: if there is a linear factor then: $p(x) = (x - a)(x^4 + bx^3 + cx^2 + dx + f)$. So: $-af = 5$. So either $a = 1, f = -5$ or $a = -1, f = 5$ or $a = 5, f = -1$ or $a = -5, f = 1$. But none of these can happen since $f(1), f(-1), f(5), f(-5)$ are all $\neq 0$ as easily checked.
case 2: if $p(x) = (x^2 + ax + b)(x^3 + cx^2 + dx + f)$. Then equating coefficients of the left and the right side we have: $bf = 5$, $af + bd = -1$, $5 + ad + bc = 3$, $d + ac + b = 2$, $a + c = 4$. There are $4$ cases again for this case:
i) $b = 1, f = 5$, then $5a + d = 1 = d + ac$. So $5a = ac$. So $a = 0$ or $c = 5$. If $a = 0$ then $d = 1$ and $d = -1$, contradiction. So $c = 5$, and $a \neq 0$.So $a = -1$. So $d = 6$ and $d = 4$, contradiction again.
ii) $b = 5, f = 1$, then: $a + 5d = -1$ and $d + ac = -3$. So: $a - 5ac = 14$ or $a(1 - 5c) = 14$. So: $a = 1, 2, 7, 14, -1, -2, -7, -14$. $a = 1, 2, 7, -2, -7, - 14$ causes $d$ to be non-integer. So $a$ can only be either $-1, 14$. If $a = -1$ then $c = 5$, but then $1 - 5c = -14$. So $c = 3$, contradiction with $c = 5$ earlier. If $a = 14$ then $1 -5c = 1$ and $c = 0$, but $c + a = 4$ so $c = 4 - 14 = -10$, contradiction again.
iii) $b = -1, f = -5$, then: $-5a - d = 1$, $ad - c = -2$, $a + c = 4$, and $ac + d = 3$. So: $ac - 5a = 2$,and this gives $a(c - 5) = 2$. So $a = 1, 2, -1, 2$. If $a = 1$ then $c = 3$ and $c = 7$, contradiction. The remaining cases also lead to contradiction of $c$ taking on different values at the same time.
iv) $b = -5, f = -1$, then: $-a - 5d = -1$, $ad - 5c = -2$, $d + ac = 7$, and $a + c = 4$. So: $-a + 5ac = 34$ or $a(5c - 1) = 34$. So $a = 1, 2, 17, 34, -1, -2, -17, -34$. None of these can happen since the system $a + c = 4$ and $a(5c - 1) = 34$ yield no integer solutions.
So in summary, the given polynomial is irreducible in $\mathbb{Q}$